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

题意：

给l到r，x到y，判断有没有两个数[l,r]/[x,y]==k的。

point：

说他水吧，死的很惨，说他不水吧，的确很水。

死在ll上面和莫名其妙的把分子分母搞反了。

不知道为什么当时考虑不到ll。

#include <iostream>#include <stdio.h>#include <string.h>#include <vector>#include <queue>using namespace std;#define LL long longint main(){    LL l, r, x, y;    LL k;    scanf("%lld %lld %lld %lld %lld", &l, &r, &x, &y, &k);    LL flag=0;    if(1.0*r/x<1.0*k||1.0*l/y>1.0*k)    {        printf("NO\n");        return 0;    }    for(LL i=x;i<=y;i++)    {        if(i*k>=l&&i*k<=r)        {            flag=1;            break;        }    }    if(flag) printf("YES\n");    else printf("NO\n");    }

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

8 477 8 1-7 3 20 2 10 -2 2-3 -3 10 6 25 3 1

output

2

input

10 840 0 90 0 101 0 11 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：

直接看图，求绿色的圆的数量。

POINT：

随便比较一下。没坑点

#include <iostream>#include <stdio.h>#include <string.h>#include <vector>#include <math.h>using namespace std;int main(){    double r,d;    cin>>r>>d;    int n;    cin>>n;    int ans=0;    for(int i=1;i<=n;i++)    {        double x,y,rr;        cin>>x>>y>>rr;        double ll=sqrt(x*x+y*y);        if(ll+rr>r) continue;        if(ll-rr<r-d) continue;        ans++;    }    printf("%d\n",ans);    }