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: Accepted)

#include<bits/stdc++.h>
using namespace std;

typedef long long int ll;
#define f(a, b) for(auto i = a;i < b;i++)

bool sortbysec(const pair<ll, ll> &a, const pair<ll,ll> &b) 
{ 
    return (a.second < b.second); 
} 
void solve()
{
	return ;
}

int main()
{
	double  sum = 0;
	ll n , m;
	cin >> n >> m;

	vector < double > vect;

	f(0, n)
	{
		double a;
		cin >> a;
		sum += a;
		vect.push_back(a);
	} 

	ll cnt = 0;
	double x = (sum/4.0)/m;
	f(0, n)
	{
		if(vect[i] >= x)
			cnt++;
	}

	if(cnt >= m)
		cout << "Yes\n";
	else
		cout << "No\n";
}