Problem:
Score : 200 points

Problem Statement

We have held a popularity poll for N items on sale. Item i received A_i votes.

From these N items, we will select M as popular items. However, we cannot select an item with less than \dfrac{1}{4M} of the total number of votes.

If M popular items can be selected, print Yes; otherwise, print No.

Constraints

1 \leq M \leq N \leq 100

1 \leq A_i \leq 1000

A_i are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 ... A_N

Output

If M popular items can be selected, print Yes; otherwise, print No.

Sample Input 1

4 1
5 4 2 1

Sample Output 1

Yes

There were 12 votes in total. The most popular item received 5 votes, and we can select it.

Sample Input 2

3 2
380 19 1

Sample Output 2

No

There were 400 votes in total. The second and third most popular items received less than \dfrac{1}{4\times 2} of the total number of votes, so we cannot select them. Thus, we cannot select two popular items.

Sample Input 3

12 3
4 56 78 901 2 345 67 890 123 45 6 789

Sample Output 3

Yes

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main(void){
    int n,m,sum=0,tmp,i,j,ans=0;
    scanf("%d%d",&n,&m);
    int a[n];
    /*入力*/
    for(i=0;n>i;++i){
        scanf("%d",&a[i]);
    }
    /*総投票数*/
    for(i=0;n>i;++i){
        sum+=a[i];
    }
    /*降順ソート*/
    for (i=0; i<n; ++i) {
        for (j=i+1; j<n; ++j) {
            if (a[i] < a[j]) {
        tmp =  a[i];
        a[i] = a[j];
        a[j] = tmp;
      }
    }
  }
  /*判定*/
  for(i=0;m>i;++i){
      if(a[i]>=(double)sum/(4*m))
      ++ans;
  }
  if(ans==m)
  printf("Yes");
  else
  printf("No");
  return 0;
}