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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define ll long long
#define INF 1<<31-1
#define LNF 9223372036854775807
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define MOD 1000000007
//#define MOD 998244353
#define NUM 200010
 
void swap(int* a,int* b){int tmp=*a;*a=*b;*b=tmp;}
int abs(int a){if(a<0)return -a;return a;}
int dup(int a,int d){if(a%d==0)return a/d;else return a/d+1;}
int mod(int a,int m){if(a<0)return a%m+m;else return a%m;}

void swapl(ll* a,ll* b){ll tmp=*a;*a=*b;*b=tmp;}
ll absl(ll a){if(a<0)return -a;return a;}
ll dupl(ll a,ll d){if(a%d==0)return a/d;else return a/d+1;}
ll modl(ll a,ll m){if(a<0)return a%m+m;else return a%m;}

ll GCD(ll a,ll b){ll c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
ll LCM(ll a,ll b){return a*b/GCD(a,b);}

ll factorial(ll n, ll m) {return n < m ? 1 : n*factorial(n-1, m);}

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

int main(){
  ll N, M;
  ll A[1000];
  ll sum=0, cnt;

  scanf("%lld%lld", &N, &M);

  for (int i = 0; i < N; i++)
  {
      scanf("%lld", A + i);
      sum += A[i];
  }
  
  for (int i = 0; i < N; i++)
  {
      if(A[i] * 4 * M >= sum) cnt++;
  }
  
  if (cnt < M)
  {
      printf("No\n");
  }
  else
  {
      printf("Yes\n");
  }
  
  return 0;
}