Educational Codeforces Round 48 (Rated for Div. 2) ABCD

A. Death Note

You received a notebook which is called Death Note. This notebook has infinite number of pages. A rule is written on the last page (huh) of this notebook. It says: "You have to write names in this notebook during $\mathrm{nn consecutive\; days.\; During\; the ii-th\; day\; you\; have\; to\; write\; exactly aiai names.".\; You\; got\; scared\; (of\; course\; you\; got\; scared,\; who\; wouldn\text{'}t\; get\; scared\; if\; he\; just\; receive\; a\; notebook\; which\; is\; named Death\; Notewith\; a\; some\; strange\; rule\; written\; in\; it?).}$

Of course, you decided to follow this rule. When you calmed down, you came up with a strategy how you will write names in the notebook. You have calculated that each page of the notebook can contain exactly $\mathrm{mm names.\; You\; will\; start\; writing\; names\; from\; the\; first\; page.\; You\; will\; write\; names\; on\; the\; current\; page\; as\; long\; as\; the\; limit\; on\; the\; number\; of\; names\; on\; this\; page\; is\; not\; exceeded.\; When\; the\; current\; page\; is\; over,\; you\; turn\; the\; page.\; Note\; that\; you always turn\; the\; page\; when\; it\; ends,\; it\; doesn\text{'}t\; matter\; if\; it\; is\; the\; last\; day\; or\; not.\; If\; after\; some\; day\; the\; current\; page\; still\; can\; hold\; at\; least\; one\; name,\; during\; the\; next\; day\; you\; will\; continue\; writing\; the\; names\; from\; the\; current\; page.}$

Now you are interested in the following question: how many times will you turn the page during each day? You are interested in the number of pages you will turn each day from $\mathrm{11 to nn.}$

Input

The first line of the input contains two integers $\mathrm{nn, mm (1\le n\le 2\cdot 1051\le n\le 2\cdot 105, 1\le m\le 1091\le m\le 109)\; —\; the\; number\; of\; days\; you\; will\; write\; names\; in\; the\; notebook\; and\; the\; number\; of\; names\; which\; can\; be\; written\; on\; each\; page\; of\; the\; notebook.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109),\; where aiai means\; the\; number\; of\; names\; you\; will\; write\; in\; the\; notebook\; during\; the ii-th\; day.}$

Output

Print exactly $\mathrm{nn integers t1,t2,\dots ,tnt1,t2,\dots ,tn,\; where titi is\; the\; number\; of\; times\; you\; will\; turn\; the\; page\; during\; the ii-th\; day.}$

Examples

input

Copy

3 5
3 7 9

output

Copy

0 2 1 

input

Copy

4 20
10 9 19 2

output

Copy

0 0 1 1 

input

Copy

1 100
99

output

Copy

0 

Note

In the first example pages of the Death Note will look like this $[1,1,1,2,2],[2,2,2,2,2],[3,3,3,3,3],[3,3,3,3][1,1,1,2,2],[2,2,2,2,2],[3,3,3,3,3],[3,3,3,3].\; Each\; number\; of\; the\; array\; describes\; during\; which\; day\; name\; on\; the\; corresponding\; position\; will\; be\; written.\; It\; is\; easy\; to\; see\; that\; you\; should\; turn\; the\; first\; and\; the\; second\; page\; during\; the\; second\; day\; and\; the\; third\; page\; during\; the\; third\; day.$

签到。、

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e6+10;
int a[N];
int main() {
    int n, m;
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    int ans = 0;
    for(int i = 1; i <= n; i ++) {
        ans += a[i];
        printf("%d ",ans/m);
        ans %= m;
    }
    return 0;
}

 

B. Segment Occurrences

You are given two strings $\mathrm{ss and tt,\; both\; consisting\; only\; of\; lowercase\; Latin\; letters.}$

The substring $\mathrm{s[l..r]s[l..r] is\; the\; string\; which\; is\; obtained\; by\; taking\; characters sl,sl+1,\dots ,srsl,sl+1,\dots ,sr without\; changing\; the\; order.}$

Each of the occurrences of string $\mathrm{aa in\; a\; string bb is\; a\; position ii (1\le i\le |b|-|a|+11\le i\le |b|-|a|+1)\; such\; that b[i..i+|a|-1]=ab[i..i+|a|-1]=a (|a||a| is\; the\; length\; of\; string aa).}$

You are asked $\mathrm{qq queries:\; for\; the ii-th\; query\; you\; are\; required\; to\; calculate\; the\; number\; of\; occurrences\; of\; string tt in\; a\; substring s[li..ri]s[li..ri].}$

Input

The first line contains three integer numbers $\mathrm{nn, mm and qq (1\le n,m\le 1031\le n,m\le 103, 1\le q\le 1051\le q\le 105)\; —\; the\; length\; of\; string ss,\; the\; length\; of\; string ttand\; the\; number\; of\; queries,\; respectively.}$

The second line is a string $\mathrm{ss (|s|=n|s|=n),\; consisting\; only\; of\; lowercase\; Latin\; letters.}$

The third line is a string $\mathrm{tt (|t|=m|t|=m),\; consisting\; only\; of\; lowercase\; Latin\; letters.}$

Each of the next $\mathrm{qq lines\; contains\; two\; integer\; numbers lili and riri (1\le li\le ri\le n1\le li\le ri\le n)\; —\; the\; arguments\; for\; the ii-th\; query.}$

Output

Print $\mathrm{qq lines\; —\; the ii-th\; line\; should\; contain\; the\; answer\; to\; the ii-th\; query,\; that\; is\; the\; number\; of\; occurrences\; of\; string tt in\; a\; substring s[li..ri]s[li..ri].}$

Examples

input

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10 3 4
codeforces
for
1 3
3 10
5 6
5 7

output

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0
1
0
1

input

Copy

15 2 3
abacabadabacaba
ba
1 15
3 4
2 14

output

Copy

4
0
3

input

Copy

3 5 2
aaa
baaab
1 3
1 1

output

Copy

0
0

Note

In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.

被坑了

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1010;
int n, m, q, l, r;
char s[N], t[N];
int a[N], cnt, sum[N];
int main() {
    cin >> n >> m >> q;
    cin >> s >> t;
    int tmp = 0;
    for(int i = 0; i < n; i ++) {
        if(s[i] == t[0]) {
            int cnt = 0;
            for(int j = 0; j < m; j ++) {
                if(s[i+j] == t[j]) cnt++;
                else break;
            }
            if(cnt == m) tmp += 1;
        }
        sum[i+1] = tmp;
    }
    // for(int i = 1; i <= n;i ++) printf("%d %d\n",i,sum[i]);
    while(q--) {
        cin >> l >> r;
        r = r-m+1;
        if(l > r) printf("0\n");
        else printf("%d\n",sum[r]-sum[l-1]);
    }
    return 0;
}

 

C. Vasya And The Mushrooms

Vasya's house is situated in a forest, and there is a mushroom glade near it. The glade consists of two rows, each of which can be divided into n consecutive cells. For each cell Vasya knows how fast the mushrooms grow in this cell (more formally, how many grams of mushrooms grow in this cell each minute). Vasya spends exactly one minute to move to some adjacent cell. Vasya cannot leave the glade. Two cells are considered adjacent if they share a common side. When Vasya enters some cell, he instantly collects all the mushrooms growing there.

Vasya begins his journey in the left upper cell. Every minute Vasya must move to some adjacent cell, he cannot wait for the mushrooms to grow. He wants to visit all the cells exactly once and maximize the total weight of the collected mushrooms. Initially, all mushrooms have a weight of 0. Note that Vasya doesn't need to return to the starting cell.

Help Vasya! Calculate the maximum total weight of mushrooms he can collect.

Input

The first line contains the number n (1 ≤ n ≤ 3·105) — the length of the glade.

The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 106) — the growth rate of mushrooms in the first row of the glade.

The third line contains n numbers b1, b2, ..., bn (1 ≤ bi ≤ 106) is the growth rate of mushrooms in the second row of the glade.

Output

Output one number — the maximum total weight of mushrooms that Vasya can collect by choosing the optimal route. Pay attention that Vasya must visit every cell of the glade exactly once.

Examples

input

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3
1 2 3
6 5 4

output

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70

input

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3
1 1000 10000
10 100 100000

output

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543210

Note

In the first test case, the optimal route is as follows:

Thus, the collected weight of mushrooms will be 0·1 + 1·2 + 2·3 + 3·4 + 4·5 + 5·6 = 70.

In the second test case, the optimal route is as follows:

Thus, the collected weight of mushrooms will be 0·1 + 1·10 + 2·100 + 3·1000 + 4·10000 + 5·100000 = 543210.

 

其实可以看成只有两种情况。

第一种：

第二种反向。

 

sum[0][i] 表示从a[0][i]出发走一圈会带a[1][i]的花费。sum[1][i] 表示a[1][i]走一圈回到a[0][i]的花费。

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 3e5+10;
ll a[2][N], s[N], sum[2][N];
int main() {
    int n;
    cin >> n;
    for(int i = 1; i <= n; i ++) scanf("%lld", &a[0][i]);
    for(int i = 1; i <= n; i ++) scanf("%lld", &a[1][i]);
    for(int i = 1; i <= n; i ++) s[i] = s[i-1] + a[0][i] + a[1][i];
    sum[0][n] = a[1][n];
    sum[1][n] = a[0][n];
    for(int i = n-1; i >= 1; i --) {
        sum[0][i] = sum[0][i+1] + s[n] - s[i] + ((n-i+1)*2-1)*a[1][i];
        sum[1][i] = sum[1][i+1] + s[n] - s[i] + ((n-i+1)*2-1)*a[0][i];
    }
    // for(int i = 0; i < 2; i ++) {
    //     for(int j = 1; j <= n; j ++) printf("%d ", sum[i][j]);
    //     printf("\n");
    // }
    ll ans = sum[0][1], cnt = 0;
    for(int i = 1; i <= n; i ++) {
        if(i&1) {
            cnt += (a[0][i]*(2*i-2)+a[1][i]*(2*i-1));
            ans = max(ans, cnt+sum[1][i+1]+(s[n]-s[i])*(2*i));
            // printf("%d %d %d %d\n",i,cnt,sum[1][i] );
        } else {
            cnt += (a[0][i]*(2*i-1)+a[1][i]*(2*i-2));
            ans = max(ans, cnt+sum[0][i+1]+(s[n]-s[i])*(2*i));
        }
    }
    cout << ans << endl;
    return 0;
}

 

D. Vasya And The Matrix

Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!

Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.

Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.

Input

The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.

The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.

The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.

Output

If there is no matrix satisfying the given constraints in the first line, output "NO".

Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.

If there are several suitable matrices, it is allowed to print any of them.

Examples

input

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2 3
2 9
5 3 13

output

Copy

YES
3 4 5
6 7 8

input

Copy

3 3
1 7 6
2 15 12

output

Copy

NO

很巧妙。

最改变第一行第一个=列的数字，然后，然后填充第一行和第一列的数字，其余全是0.

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 110;
ll n, m;
ll x[N], y[N];
ll a[N][N];
int main() {
    cin >> n >> m;
    ll ansx = 0;
    for(int i = 1; i <= n; i ++) {
        cin >> x[i];
        ansx ^= x[i];
    }
    ll ansy = 0;
    for(int i = 1; i <= m; i ++) cin >> y[i], ansy ^= y[i];
    if(ansy == ansx) printf("YES\n");
    else return 0*printf("NO\n");
    ansx ^= x[1];
    ansx ^= y[1];
    a[1][1] = ansx;
    for(int i = 2; i <= m; i ++) a[1][i] = y[i];
    for(int i = 2; i <= n; i ++) a[i][1] = x[i];
    for(int i = 1; i <= n; i ++) {
        for(int j = 1 ; j <= m; j ++) {
            printf("%d ",a[i][j]);
        }printf("\n");
    }
    return 0;
}