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 in;
typedef pair<in,in> ii;
#define F first
#define S second
#define sqr(x) (x)*(x)
#define pb(x) push_back(x)
#define sz(x) (in)x.size()
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define rep(i,x,y) for(in i=x;i<(y);++i)
#define IOS ios_base::sync_with_stdio(0);cin.tie(0)
/* start */

int main()
{
    IOS;
    in n,m,tot=0;
    cin >> n >> m;
    vector<in> p(n);
    rep(i,0,n){
        cin >> p[i];
        tot+=p[i];
    }
    in k = tot/(4*m),count=0;
    if(tot % (4*m) != 0) k++;
    rep(i,0,n){
        if(p[i]>=k) count++;
    }
    (count>=m) ? cout << "Yes\n" : cout << "No\n";
}