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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// code.cpp : This file contains the 'main' function. Program execution begins and ends there.
//

#include <bits/stdc++.h>
#define ll long long

using namespace std;

int main()
{
int n,m,A;
A=0;
cin>>n>>m;
int a[n];
for(int i=0; i<n; i++){
	cin>>a[i];
}
int sum =0;
for(int i=0; i<n; i++){
	sum = sum+a[i];
}
int p = 4*m;
int val = sum/p;
for(int i=0; i<n; i++){
	if(a[i]>=val){
		A++;
	}
}
if(A>=m){
	cout<<"Yes";
}
else{
	cout<<"No";
}

}