Codeforces Round #430 (Div. 2)

A. Kirill And The Game

Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number.

For each two integer numbers a and b such that l ≤ a ≤ r and x ≤ b ≤ y there is a potion with experience a and cost b in the store (that is, there are (r - l + 1)·(y - x + 1) potions).

Kirill wants to buy a potion which has efficiency k. Will he be able to do this?

Input

First string contains five integer numbers l, r, x, y, k (1 ≤ l ≤ r ≤ 107, 1 ≤ x ≤ y ≤ 107, 1 ≤ k ≤ 107).

Output

Print "YES" without quotes if a potion with efficiency exactly k can be bought in the store and "NO" without quotes otherwise.

You can output each of the letters in any register.

Examples

input

1 10 1 10 1

output

YES

input

1 5 6 10 1

output

NO

 

题意：[l,r],[x,y] 两个区间的数能否组成 k。

枚举一个变量，另一个变量查看。

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int l,r,x,y,k;
    cin>>l>>r>>x>>y>>k;

    bool flag = false;
    for(int i=l; i <= r; i++)
        if(i%k==0&&i/k>=x&&i/k<=y)
        {
            flag = true;
            break;
        }

    if(flag)
        puts("YES");
    else puts("NO");


    return 0;
}

 

B. Gleb And Pizza

Gleb ordered pizza home. When the courier delivered the pizza, he was very upset, because several pieces of sausage lay on the crust, and he does not really like the crust.

The pizza is a circle of radius r and center at the origin. Pizza consists of the main part — circle of radius r - d with center at the origin, and crust around the main part of the width d. Pieces of sausage are also circles. The radius of the i -th piece of the sausage is ri, and the center is given as a pair (xi, yi).

Gleb asks you to help determine the number of pieces of sausage caught on the crust. A piece of sausage got on the crust, if it completely lies on the crust.

Input

First string contains two integer numbers r and d (0 ≤ d < r ≤ 500) — the radius of pizza and the width of crust.

Next line contains one integer number n — the number of pieces of sausage (1 ≤ n ≤ 105).

Each of next n lines contains three integer numbers xi, yi and ri ( - 500 ≤ xi, yi ≤ 500, 0 ≤ ri ≤ 500), where xi and yi are coordinates of the center of i-th peace of sausage, ri — radius of i-th peace of sausage.

Output

Output the number of pieces of sausage that lay on the crust.

Examples

input

8 4
7
7 8 1
-7 3 2
0 2 1
0 -2 2
-3 -3 1
0 6 2
5 3 1

output

2

input

10 8
4
0 0 9
0 0 10
1 0 1
1 0 2

output

0

Note

Below is a picture explaining the first example. Circles of green color denote pieces of sausage lying on the crust.

题意：查看绿色的圆的个数。

分析：半径关系。

#include <bits/stdc++.h>

using namespace std;

int main()
{
    double r,d;
    cin>>r>>d;

    int n;
    scanf("%d",&n);
    int cnt = 0;
    while(n--) {
        double x,y,m;
        cin>>x>>y>>m;

        if( sqrt(x*x+y*y) - m  >= r-d && sqrt(x*x+y*y)+m <=r)
            cnt++;
    }

    cout << cnt <<endl;

    return 0;
}

 

C. Ilya And The Tree

Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex 1. There is an integer number written on each vertex of the tree; the number written on vertex i is equal to ai.

Ilya believes that the beauty of the vertex x is the greatest common divisor of all numbers written on the vertices on the path from the root to x, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to 0 or leave all vertices unchanged. Now for each vertex Ilya wants to know the maximum possible beauty it can have.

For each vertex the answer must be considered independently.

The beauty of the root equals to number written on it.

Input

First line contains one integer number n — the number of vertices in tree (1 ≤ n ≤ 2·105).

Next line contains n integer numbers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 2·105).

Each of next n - 1 lines contains two integer numbers x and y (1 ≤ x, y ≤ n, x ≠ y), which means that there is an edge (x, y) in the tree.

Output

Output n numbers separated by spaces, where i-th number equals to maximum possible beauty of vertex i.

Examples

input

2
6 2
1 2

output

6 6 

input

3
6 2 3
1 2
1 3

output

6 6 6 

input

1
10

output

10 

题意：求一个结点到根的路径中，可以删掉一个点，这些点的最大的 gcd 是多少。

分析：暴力，这个DFS写的很巧妙，传了一个set的参数，表示这条路上删掉某一个点的gcd和总和的集合。

参考：http://blog.csdn.net/harlow_cheng/article/details/77692767

#include <bits/stdc++.h>

using namespace std;


const int maxn = 200005;
vector<int> g[maxn];

int n;
int a[maxn];

int gcd (int a,int b) {
    return b == 0 ? a : gcd(b,a%b);
}


int ans[maxn],now[maxn];

void dfs(int u,int fa,set<int> s) {

    for(int i=0; i < (int)g[u].size(); i++) {
        int v = g[u][i];
        if(v==fa) continue;
        now[v] = gcd(now[u],a[v]);
        set<int> tset;
        set<int> :: iterator it;

        for(it=s.begin();it!=s.end();it++) {
            int tmp = *it;
            int tmp2 = gcd(tmp,a[v]);
            tset.insert(gcd(tmp,a[v]));
        }
        tset.insert(now[u]);
        ans[v] = max(now[v],*(--tset.end()));
        dfs(v,u,tset);
    }

}

int main()
{
    freopen("in.txt","r",stdin);
    scanf("%d",&n);

    for(int i = 1; i <= n; i++)
        scanf("%d",&a[i]);

    for(int i = 1; i < n; i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        g[v].push_back(u);
    }

    ans[1] = now[1] = a[1];
    dfs(1,0,{0});

    for(int i = 1; i <= n; i++)
        printf("%d ",ans[i]);

    return 0;
}