Codeforces Round #481 (Div. 3） ABCDEFG

A. Remove Duplicates

Petya has an array $\mathrm{aa consisting\; of nn integers.\; He\; wants\; to\; remove\; duplicate\; (equal)\; elements.}$

Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 50)\; —\; the\; number\; of\; elements\; in\; Petya\text{'}s\; array.}$

The following line contains a sequence${\mathrm{a1,a2,\dots ,an (1\le ai\le 10001\le ai\le 1000)\; —\; the\; Petya\text{'}s\; array.}}^{}$

Output

In the first line print integer $\mathrm{xx —\; the\; number\; of\; elements\; which\; will\; be\; left\; in\; Petya\text{'}s\; array\; after\; he\; removed\; the\; duplicates.}$

In the second line print $\mathrm{xx integers\; separated\; with\; a\; space\; —\; Petya\text{'}s\; array\; after\; he\; removed\; the\; duplicates.\; For\; each\; unique\; element\; only\; the\; rightmost\; entry\; should\; be\; left.}$

Examples

input

Copy

6
1 5 5 1 6 1

output

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3
5 6 1 

input

Copy

5
2 4 2 4 4

output

Copy

2
2 4 

input

Copy

5
6 6 6 6 6

output

Copy

1
6 

Note

In the first example you should remove two integers $\mathrm{11,\; which\; are\; in\; the\; positions 11 and 44.\; Also\; you\; should\; remove\; the\; integer 55,\; which\; is\; in\; the\; position 22.}$

In the second example you should remove integer $\mathrm{22,\; which\; is\; in\; the\; position 11,\; and\; two\; integers 44,\; which\; are\; in\; the\; positions 22 and 44.}$

In the third example you should remove four integers $\mathrm{66,\; which\; are\; in\; the\; positions 11, 22, 33 and 44.}$

 

 去掉重复的，留下重复中最后一个数。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int a[110];
map<int,int> mp;
vector<int> vs;
int main() {
    int n, x;
    cin >> n;
    for(int i = 0; i < n; i ++) {
        cin >> a[i];
    }
    for(int i = n-1; i >= 0; i --) {
        if(!mp[a[i]]) {
            vs.push_back(a[i]);
            mp[a[i]] = 1;
        }
    }
    cout << vs.size() << endl;
    int len = vs.size();
    for(int i = len-1; i >= 0; i --) printf("%d ",vs[i]);
    printf("\n");
    return 0;
}

B. File Name

You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.

Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".

You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $\mathrm{11.\; For\; example,\; if\; you\; delete\; the\; character\; in\; the\; position 22 from\; the\; string\; "exxxii",\; then\; the\; resulting\; string\; is\; "exxii".}$

Input

The first line contains integer $\mathrm{nn (3\le n\le 100) —\; the\; length\; of\; the\; file\; name.}$

The second line contains a string of length $\mathrm{nn consisting\; of\; lowercase\; Latin\; letters\; only\; —\; the\; file\; name.}$

Output

Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.

Examples

input

Copy

6
xxxiii

output

Copy

1

input

Copy

5
xxoxx

output

Copy

0

input

Copy

10
xxxxxxxxxx

output

Copy

8

Note

In the first example Polycarp tried to send a file with name contains number $\mathrm{3333,\; written\; in\; Roman\; numerals.\; But\; he\; can\; not\; just\; send\; the\; file,\; because\; it\; name\; contains\; three\; letters\; "x"\; in\; a\; row.\; To\; send\; the\; file\; he\; needs\; to\; remove\; any\; one\; of\; this\; letters.}$

有三个或以上的‘x’字符，就删除某些字符使的最多只有两个‘x’字符。问最少删除多少个。

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
    int n;
    string s;
    cin >> n >> s;
    int ans = 0;
    for(int i = 0; i < n; i ++) {
        int cnt = 0;
        while(s[i] =='x') {
            cnt++;
            i++;
        }
        if(cnt >= 3)ans += cnt-2;
    }
    cout << ans << endl;
    return 0;
}

C. Letters

There are $\mathrm{nn dormitories\; in\; Berland\; State\; University,\; they\; are\; numbered\; with\; integers\; from 11 to nn.\; Each\; dormitory\; consists\; of\; rooms,\; there\; are aiai rooms\; in ii-th\; dormitory.\; The\; rooms\; in ii-th\; dormitory\; are\; numbered\; from 11 to aiai.}$

A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $\mathrm{nn dormitories\; is\; written\; on\; an\; envelope.\; In\; this\; case,\; assume\; that\; all\; the\; rooms\; are\; numbered\; from 11 to a1+a2+\cdots +ana1+a2+\cdots +an and\; the\; rooms\; of\; the\; first\; dormitory\; go\; first,\; the\; rooms\; of\; the\; second\; dormitory\; go\; after\; them\; and\; so\; on.}$

For example, in case $\mathrm{n=2n=2, a1=3a1=3 and a2=5a2=5 an\; envelope\; can\; have\; any\; integer\; from 11 to 88 written\; on\; it.\; If\; the\; number 77 is\; written\; on\; an\; envelope,\; it\; means\; that\; the\; letter\; should\; be\; delivered\; to\; the\; room\; number 44 of\; the\; second\; dormitory.}$

For each of $\mathrm{mm letters\; by\; the\; room\; number\; among\; all nn dormitories,\; determine\; the\; particular\; dormitory\; and\; the\; room\; number\; in\; a\; dormitory\; where\; this\; letter\; should\; be\; delivered.}$

Input

The first line contains two integers $\mathrm{nn and mm (1\le n,m\le 2\cdot 105) —\; the\; number\; of\; dormitories\; and\; the\; number\; of\; letters.}$

The second line contains a sequence ${\mathrm{a1,a2,\dots ,an(1\le ai\le 1010),\; where aiai equals\; to\; the\; number\; of\; rooms\; in\; the ii-th\; dormitory.\; The\; third\; line\; contains\; a\; sequence b1,b2,\dots ,bm (1\le bj\le a1+a2+\cdots +an)(1\le bj\le a1+a2+\cdots +an),\; where bjbj equals\; to\; the\; room\; number\; (among\; all\; rooms\; of\; all\; dormitories)\; for\; the jj-th\; letter.\; All bjbj are\; given\; in increasing order.}}^{}$

Output

Print $\mathrm{mm lines.\; For\; each\; letter\; print\; two\; integers f and k —\; the\; dormitory\; number f(1\le f\le n) and\; the\; room\; number kk in\; this\; dormitory (1\le k\le af)(1\le k\le af) to\; deliver\; the\; letter.}$

Examples

input

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3 6
10 15 12
1 9 12 23 26 37

output

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1 1
1 9
2 2
2 13
3 1
3 12

input

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2 3
5 10000000000
5 6 9999999999

output

Copy

1 5
2 1
2 9999999994

Note

In the first example letters should be delivered in the following order:

• the first letter in room $\mathrm{11 of\; the\; first\; dormitory}$
• the second letter in room $\mathrm{99 of\; the\; first\; dormitory}$
• the third letter in room $\mathrm{22 of\; the\; second\; dormitory}$
• the fourth letter in room $\mathrm{1313 of\; the\; second\; dormitory}$
• the fifth letter in room $\mathrm{11 of\; the\; third\; dormitory}$
• the sixth letter in room $\mathrm{1212 of\; the\; third\; dormitory}$

$\mathrm{有3栋楼，每楼有ai个，现在从第一栋开始标记为1，知道a1+a2+...+an，}$

$\mathrm{问给定一个数k，求现在在第ji栋，第几号房。二分查找。}$

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
ll a[N], b[N];
int main() {
    ll n, m;
    cin >> n >> m >> a[0];
    b[0] = a[0];
    for(int i = 1; i < n; i ++) {
        cin >> a[i];
        b[i] = b[i-1] + a[i];
    }
    while(m--) {
        ll x;
        cin >> x;
        int id = lower_bound(b,b+n,x) -b;
        printf("%d %lld\n",id+1,a[id]-(b[id]-x));
    }
    return 0;
}

D. Almost Arithmetic Progression

Polycarp likes arithmetic progressions. A sequence $[a1,a2,\dots ,an][a1,a2,\dots ,an] is\; called\; an\; arithmetic\; progression\; if\; for\; each ii (1\le i<n1\le i<n)\; the\; value ai+1-aiai+1-ai is\; the\; same.\; For\; example,\; the\; sequences [42][42], [5,5,5][5,5,5], [2,11,20,29][2,11,20,29] and [3,2,1,0][3,2,1,0] are\; arithmetic\; progressions,\; but [1,0,1][1,0,1], [1,3,9][1,3,9] and [2,3,1][2,3,1] are\; not.$

It follows from the definition that any sequence of length one or two is an arithmetic progression.

Polycarp found some sequence of positive integers $[b1,b2,\dots ,bn][b1,b2,\dots ,bn].\; He\; agrees\; to\; change\; each\; element\; by\; at\; most\; one.\; In\; the\; other\; words,\; for\; each\; element\; there\; are\; exactly\; three\; options:\; an\; element\; can\; be\; decreased\; by 11,\; an\; element\; can\; be\; increased\; by 11,\; an\; element\; can\; be\; left\; unchanged.$

Determine a minimum possible number of elements in $\mathrm{bb which\; can\; be\; changed\; (by\; exactly\; one),\; so\; that\; the\; sequence bb becomes\; an\; arithmetic\; progression,\; or\; report\; that\; it\; is\; impossible.}$

It is possible that the resulting sequence contains element equals $00.$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 100000)(1\le n\le 100000) —\; the\; number\; of\; elements\; in bb.}$

The second line contains a sequence ${\mathrm{b1,b2,\dots ,bnb1,b2,\dots ,bn (1\le bi\le 109)(1\le bi\le 109).}}^{}$

Output

If it is impossible to make an arithmetic progression with described operations, print -1. In the other case, print non-negative integer — the minimum number of elements to change to make the given sequence becomes an arithmetic progression. The only allowed operation is to add/to subtract one from an element (can't use operation twice to the same position).

Examples

input

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4
24 21 14 10

output

Copy

3

input

Copy

2
500 500

output

Copy

0

input

Copy

3
14 5 1

output

Copy

-1

input

Copy

5
1 3 6 9 12

output

Copy

1

Note

In the first example Polycarp should increase the first number on $\mathrm{11,\; decrease\; the\; second\; number\; on 11,\; increase\; the\; third\; number\; on 11,\; and\; the\; fourth\; number\; should\; left\; unchanged.\; So,\; after\; Polycarp\; changed\; three\; elements\; by\; one,\; his\; sequence\; became\; equals\; to [25,20,15,10][25,20,15,10],\; which\; is\; an\; arithmetic\; progression.}$

In the second example Polycarp should not change anything, because his sequence is an arithmetic progression.

In the third example it is impossible to make an arithmetic progression.

In the fourth example Polycarp should change only the first element, he should decrease it on one. After that his sequence will looks like $[0,3,6,9,12][0,3,6,9,12],\; which\; is\; an\; arithmetic\; progression.$

每个数可以加1、减1,、不变，问最少改变多少次使得数列变成等差数列，否则是-1。有等差数有9种情况。每一中都试下，取可以构成等差数列的最小值。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
int a[N], b[N];
int dx[] = {1, 1, 1, 0, 0, 0, -1, -1, -1},
    dy[] = {1, 0, -1, 1, 0, -1, 1, 0, -1};
int main() {
    int n, s = 0;
    cin >> n;
    for(int i = 0; i < n; i ++) {
        cin >> a[i];
    }
    if(n <= 2) return 0*printf("0\n");
    int ans = 1e7;
    for(int i = 0; i < 9; i ++) {
        int cnt = 0, flag = 1;
        if(dx[i]) cnt++;
        if(dy[i]) cnt++;
        b[0] = a[0] + dx[i];
        b[1] = a[1] + dy[i];
        int tmp = b[1] - b[0];
        for(int j = 2; j < n; j ++) {
            if(a[j]+1-b[j-1] == tmp) {
                b[j] = a[j] + 1;
                cnt++;
            } else if(a[j] - b[j-1] == tmp) b[j] = a[j];
            else if(a[j]-1-b[j-1] == tmp) {
                b[j] = a[j]-1;
                cnt++;
            } else {
                flag = 0;
                break;
            }
        }
        if(flag) ans = min(ans, cnt);
    }
    printf("%d\n",ans==1e7?-1:ans);
    return 0;
}

E. Bus Video System

The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.

If $\mathrm{xx is\; the\; number\; of\; passengers\; in\; a\; bus\; just\; before\; the\; current\; bus\; stop\; and yy is\; the\; number\; of\; passengers\; in\; the\; bus\; just\; after\; current\; bus\; stop,\; the\; system\; records\; the\; number y-xy-x.\; So\; the\; system\; records\; show\; how\; number\; of\; passengers\; changed.}$

The test run was made for single bus and $\mathrm{nn bus\; stops.\; Thus,\; the\; system\; recorded\; the\; sequence\; of\; integers a1,a2,\dots ,ana1,a2,\dots ,an (exactly\; one\; number\; for\; each\; bus\; stop),\; where aiai is\; the\; record\; for\; the\; bus\; stop ii.\; The\; bus\; stops\; are\; numbered\; from 11 to nn in\; chronological\; order.}$

Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $\mathrm{ww (that\; is,\; at\; any\; time\; in\; the\; bus\; there\; should\; be\; from 00 to ww passengers\; inclusive).}$

Input

The first line contains two integers $\mathrm{nn and ww (1\le n\le 1000,1\le w\le 109)(1\le n\le 1000,1\le w\le 109) —\; the\; number\; of\; bus\; stops\; and\; the\; capacity\; of\; the\; bus.}$

The second line contains a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (-106\le ai\le 106)(-106\le ai\le 106),\; where aiai equals\; to\; the\; number,\; which\; has\; been\; recorded\; by\; the\; video\; system\; after\; the ii-th\; bus\; stop.}}^{}$

Output

Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to $\mathrm{ww.\; If\; the\; situation\; is\; contradictory\; (i.e.\; for\; any\; initial\; number\; of\; passengers\; there\; will\; be\; a\; contradiction),\; print 0.}$

Examples

input

Copy

3 5
2 1 -3

output

Copy

3

input

Copy

2 4
-1 1

output

Copy

4

input

Copy

4 10
2 4 1 2

output

Copy

2

Note

In the first example initially in the bus could be $\mathrm{00, 11 or 22 passengers.}$

In the second example initially in the bus could be $\mathrm{11, 22, 33 or 44 passengers.}$

In the third example initially in the bus could be $\mathrm{00 or 11 passenger.}$

 

 问第1个车站之前车上有多少人，共有多少种情况，不成立就是0。先假设之前有0个人，然后前缀和，取最大值和最小值，小于0的话明天0不成立，这就要算需要加多少使的都不小于0了。然后w-MAX+1，都小于0特判下。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1010;
ll n, w, a[N];
ll max(ll x, ll y) {
    return x > y ? x: y;
}
int main() {
    cin >> n >> w >> a[0];
    for(int i = 1; i < n; i ++) {
        cin >> a[i];
        a[i] += a[i-1];
    }
    ll MIN = 1e12, MAX = -1e12;
    for(int i = 0; i < n; i ++) {
        MIN = min(MIN, a[i]);
        MAX = max(MAX, a[i]);
    }
    // cout << MIN << ' ' << MAX << endl;
    if(MIN < 0 && MAX < 0) {
        cout << max(0,w + MIN+1) << endl;
    } else if(MAX >= 0 && MIN < 0) {
        MAX -= MIN;
        cout << max(0,w - MAX + 1) << endl;
    } else if(MAX >= 0 && MIN >= 0) {
        cout << max(0, w - MAX + 1) << endl;
    }
    return 0;
}

F. Mentors

In BerSoft $\mathrm{nn programmers\; work,\; the\; programmer ii is\; characterized\; by\; a\; skill riri.}$

A programmer $\mathrm{aa can\; be\; a\; mentor\; of\; a\; programmer bb if\; and\; only\; if\; the\; skill\; of\; the\; programmer aa is\; strictly\; greater\; than\; the\; skill\; of\; the\; programmer bb (ra>rb)(ra>rb) and\; programmers aa and bb are\; not\; in\; a\; quarrel.}$

You are given the skills of each programmers and a list of $\mathrm{kk pairs\; of\; the\; programmers,\; which\; are\; in\; a\; quarrel\; (pairs\; are\; unordered).\; For\; each\; programmer ii,\; find\; the\; number\; of\; programmers,\; for\; which\; the\; programmer ii can\; be\; a\; mentor.}$

Input

The first line contains two integers $\mathrm{nn and kk (2\le n\le 2\cdot 105(2\le n\le 2\cdot 105, 0\le k\le min(2\cdot 105,n\cdot (n-1)2))0\le k\le min(2\cdot 105,n\cdot (n-1)2)) —\; total\; number\; of\; programmers\; and\; number\; of\; pairs\; of\; programmers\; which\; are\; in\; a\; quarrel.}$

The second line contains a sequence of integers ${\mathrm{r1,r2,\dots ,rnr1,r2,\dots ,rn (1\le ri\le 109)(1\le ri\le 109),\; where riri equals\; to\; the\; skill\; of\; the ii-th\; programmer.}}^{}$

Each of the following $\mathrm{kk lines\; contains\; two\; distinct\; integers xx, yy (1\le x,y\le n(1\le x,y\le n, x\ne y)x\ne y) —\; pair\; of\; programmers\; in\; a\; quarrel.\; The\; pairs\; are\; unordered,\; it\; means\; that\; if xx is\; in\; a\; quarrel\; with yy then yy is\; in\; a\; quarrel\; with xx.\; Guaranteed,\; that\; for\; each\; pair (x,y)(x,y) there\; are\; no\; other\; pairs (x,y)(x,y) and (y,x)(y,x) in\; the\; input.}$

Output

Print $\mathrm{nn integers,\; the ii-th\; number\; should\; be\; equal\; to\; the\; number\; of\; programmers,\; for\; which\; the ii-th\; programmer\; can\; be\; a\; mentor.\; Programmers\; are\; numbered\; in\; the\; same\; order\; that\; their\; skills\; are\; given\; in\; the\; input.}$

Examples

input

Copy

4 2
10 4 10 15
1 2
4 3

output

Copy

0 0 1 2 

input

Copy

10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5

output

Copy

5 4 0 5 3 3 9 0 2 5 

Note

In the first example, the first programmer can not be mentor of any other (because only the second programmer has a skill, lower than first programmer skill, but they are in a quarrel). The second programmer can not be mentor of any other programmer, because his skill is minimal among others. The third programmer can be a mentor of the second programmer. The fourth programmer can be a mentor of the first and of the second programmers. He can not be a mentor of the third programmer, because they are in a quarrel.

求小于ai的数有多少，并且减去和i有争吵的。

先排序，二分查找小于ai的数有多少个，然后看与 i 有争吵的数是否小于ai，小于的话减一。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
int a[N], b[N], n, k;
vector<int> vs[N];
int main() {
    cin >> n >> k;
    for(int i = 0; i < n; i ++) {
        cin >> a[i];
        b[i] = a[i];
    }
    sort(b,b+n);
    for(int i = 1, u, v; i <= k; i ++) {
        cin >> u >> v;
        vs[u-1].push_back(v-1);
        vs[v-1].push_back(u-1);
    }
    for(int i = 0; i < n; i ++) {
        int id = lower_bound(b,b+n,a[i])-b;
        for(int j = 0; j < vs[i].size(); j ++) {
            if(a[vs[i][j]] < a[i]) id--;
        }
        printf("%d ",id);
    }
    printf("\n");
    return 0;
}

G. Petya's Exams

Petya studies at university. The current academic year finishes with $\mathrm{nn special\; days.\; Petya\; needs\; to\; pass mm exams\; in\; those\; special\; days.\; The\; special\; days\; in\; this\; problem\; are\; numbered\; from 11 to nn.}$

There are three values about each exam:

• ${\mathrm{sisi —\; the\; day,\; when\; questions\; for\; the ii-th\; exam\; will\; be\; published,}}^{}$
• ${\mathrm{didi —\; the\; day\; of\; the ii-th\; exam\; (si<disi<di),}}^{}$
• ${\mathrm{cici —\; number\; of\; days\; Petya\; needs\; to\; prepare\; for\; the ii-th\; exam.\; For\; the ii-th\; exam\; Petya\; should\; prepare\; in\; days\; between sisi and di-1di-1,\; inclusive.}}^{}$

There are three types of activities for Petya in each day: to spend a day doing nothing (taking a rest), to spend a day passing exactly one exam or to spend a day preparing for exactly one exam. So he can't pass/prepare for multiple exams in a day. He can't mix his activities in a day. If he is preparing for the $\mathrm{ii-th\; exam\; in\; day jj,\; then si\le j<disi\le j<di.}$

It is allowed to have breaks in a preparation to an exam and to alternate preparations for different exams in consecutive days. So preparation for an exam is not required to be done in consecutive days.

Find the schedule for Petya to prepare for all exams and pass them, or report that it is impossible.

Input

The first line contains two integers $\mathrm{nn and mm (2\le n\le 100,1\le m\le n)(2\le n\le 100,1\le m\le n) —\; the\; number\; of\; days\; and\; the\; number\; of\; exams.}$

Each of the following $\mathrm{mm lines\; contains\; three\; integers sisi, didi, cici (1\le si<di\le n,1\le ci\le n)(1\le si<di\le n,1\le ci\le n) —\; the\; day,\; when\; questions\; for\; the ii-th\; exam\; will\; be\; given,\; the\; day\; of\; the ii-th\; exam,\; number\; of\; days\; Petya\; needs\; to\; prepare\; for\; the ii-th\; exam.}$

Guaranteed, that all the exams will be in different days. Questions for different exams can be given in the same day. It is possible that, in the day of some exam, the questions for other exams are given.

Output

If Petya can not prepare and pass all the exams, print -1. In case of positive answer, print $\mathrm{nn integers,\; where\; the jj-th\; number\; is:}$

• $(m+1)(m+1),\; if\; the jj-th\; day\; is\; a\; day\; of\; some\; exam\; (recall\; that\; in\; each\; day\; no\; more\; than\; one\; exam\; is\; conducted),$
• zero, if in the $\mathrm{jj-th\; day\; Petya\; will\; have\; a\; rest,}$
• $\mathrm{ii (1\le i\le m1\le i\le m),\; if\; Petya\; will\; prepare\; for\; the ii-th\; exam\; in\; the\; day jj (the\; total\; number\; of\; days\; Petya\; prepares\; for\; each\; exam\; should\; be strictly equal\; to\; the\; number\; of\; days\; needed\; to\; prepare\; for\; it).}$

  Assume that the exams are numbered in order of appearing in the input, starting from $11.$

  If there are multiple schedules, print any of them.

Examples

input

Copy

5 2
1 3 1
1 5 1

output

Copy

1 2 3 0 3 

input

Copy

3 2
1 3 1
1 2 1

output

Copy

-1

input

Copy

10 3
4 7 2
1 10 3
8 9 1

output

Copy

2 2 2 1 1 0 4 3 4 4 

Note

In the first example Petya can, for example, prepare for exam $\mathrm{11 in\; the\; first\; day,\; prepare\; for\; exam 22 in\; the\; second\; day,\; pass\; exam 11 in\; the\; third\; day,\; relax\; in\; the\; fourth\; day,\; and\; pass\; exam 22 in\; the\; fifth\; day.\; So,\; he\; can\; prepare\; and\; pass\; all\; exams.}$

In the second example, there are three days and two exams. So, Petya can prepare in only one day (because in two other days he should pass exams). Then Petya can not prepare and pass all exams.

贪心问题。先让di考试这天全标记为m+1，然后按考试时间排序。对于每一门考试，从[si,di-1]遍历ci次。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 110;
struct Nod{
    int a, b, c, id;
}nod[N];
bool cmp(Nod &a, Nod &b) {
    return a.b < b.b;
}
int b[N];
int main() {
    int n, m;
    cin >> n >> m;
    for(int i = 0; i < m; i ++) {
        cin >> nod[i].a >> nod[i].b >> nod[i].c;
        if(b[nod[i].b] != 0) return 0*printf("-1\n");
        b[nod[i].b] = m+1;
        nod[i].id = i+1;
    }
    for(int i = 0; i < m; i ++) {
        int ans = 0;
        for(int j = nod[i].a; j < nod[i].b; j++) {
            if(b[j] == 0) {
                b[j] = nod[i].id;
                ans++;
            }
            if(ans == nod[i].c) break;
        }
        if(ans < nod[i].c) return 0*printf("-1\n");
    }
    for(int i = 1; i <= n; i ++) printf("%d ",b[i]);
    printf("\n");
    return 0;
}