Codeforces Round #430 (Div. 2)

A. Kirill And The Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

 水题

#include<bits/stdc++.h>
#define pb push_back
#define ll long long
using namespace std;
const int maxn=1e5+5;
int l,r,x,y,k;
vector<int>s;
int main()
{    
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    cin>>l>>r>>x>>y>>k;
    //int a=floor((float)l/y)+1,b=floor((float)r/x);
    
    for(int i=x;i<=y;i++)
    {
        if((ll)i*k>=l&&(ll)i*k<=r)
        {
            puts("YES");return 0;
        }
    }
        puts("NO");
    return 0;
}

B. Gleb And Pizza

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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：问落在外环的有多少个园。

#include<bits/stdc++.h>
#define pb push_back
#define ll long long
using namespace std;
const int maxn=1e5+5;
int d,R,n;
double len(double x,double y)
{
    return sqrt(x*x+y*y);
}
int main()
{    
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    cin>>R>>d>>n;
    int ans=0;
    while(n--)
    {
        double x,y,r;
        cin>>x>>y>>r;
        if(2*r<=d)
        {
            double tmp=len(x,y);
            if(tmp-r>=R-d&&tmp+r<=R)
            {
                ans++;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

C. Ilya And The Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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：给一棵树，根节点为1，每一个节点有一个值，对与每一个节点来说他的漂亮值为根节点到此节点的所有值的gcd,但可以把其中一个值改为0；

题解：用b[i]记录从根节点到此节点i的所有值的gcd.用set,记录到某个节点，以前有一个节点改为0的所有值的gcd,然后dfs去维护set,和b[]，每一个答案就是set和b[i]中的较大值；

#include<bits/stdc++.h>
#define pb push_back
#define ll long long
using namespace std;
const int maxn=2e5+5;
int n,a[maxn],x,m;
vector<int>edg[maxn];
set<int>cnt[maxn];
int b[maxn];
bool vis[maxn];
void dfs(int v,int f)
{
    vis[v]=true;
    int len=edg[v].size(); 
    set<int>::iterator it;
    b[v]=__gcd(a[v],b[f]);
    for(it=cnt[f].begin();it!=cnt[f].end();it++)
    {
        cnt[v].insert(__gcd(a[v],*it));
    }
    cnt[v].insert(b[f]);
    for(int i=0;i<len;i++)
    {
        if(!vis[edg[v][i]])dfs(edg[v][i],v);
    }
    return ;
}
int main()
{    
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        edg[x].pb(y);
        edg[y].pb(x);
    }
    dfs(1,0);
    set<int>::iterator it;
    for(int i=1;i<=n;i++)
    {
        it=--cnt[i].end(); 
        int ans=*it;
        ans=max(ans,b[i]);
        cout<<ans<<' ';
    }
    return 0;
}
close