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>
#define all(x) (x).begin(), (x).end()
#define gc getchar_unlocked()
#define pc(x) putchar_unlocked(x)
template<typename T> void scan(T &x){x = 0;bool _=0;T c=gc;_=c==45;c=_?gc:c;while(c<48||c>57)c=gc;for(;c<48||c>57;c=gc);for(;c>47&&c<58;c=gc)x=(x<<3)+(x<<1)+(c&15);x=_?-x:x;}
template<typename T> void printn(T n){bool _=0;_=n<0;n=_?-n:n;char snum[65];int i=0;do{snum[i++]=n%10+48;n/= 10;}while(n);--i;if (_)pc(45);while(i>=0)pc(snum[i--]);}
template<typename First, typename ... Ints> void scan(First &arg, Ints&... rest){scan(arg);scan(rest...);}
template<typename T> void print(T n){printn(n);pc(10);}
template<typename First, typename ... Ints> void print(First arg, Ints... rest){printn(arg);pc(32);print(rest...);}

using namespace std;

int n, m, a[1002], sum;

int main(){
    scan(n, m);
    for(int i = 0; i < n; i++){
        scan(a[i]);
        sum += a[i];
    }
    sum = ceil(1.0*sum/(4.0*m));
    int cnt = 0;
    for(int i = 0; i < n ;i++){
        if(a[i] >= sum)
            cnt++;
    }
    
    if(cnt >= m)
        puts("Yes");
    else
        puts("No");
    
    return 0;
}