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 9.2.1) (judge verdict: Accepted)

/*
ID: learnin7
TASK: test
LANG: C++                 
*/
/* LANG can be C++11 or C++14 for those more recent releases */
#include<bits/stdc++.h>

/*  Macros from Endagorion */
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

using namespace std;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;


int main(){
        
    ios_base::sync_with_stdio(0);
    
    //ofstream fout ("test.out");
    //ifstream fin ("test.in");
    //fin >> a >> b;
    //fout << a+b << endl;
    int n,m;
    cin>>n>>m;
    int x[n];
    int tot=0;
    forn(i,n){
        cin>>x[i];
        tot += x[i];
    }
    sort(x,x+n);
    bool flag = true;
    for(int i=n-1;i>=n-m;i--){
        if(4*m*x[i]<tot){
            flag = false;
            break;
        }
    }
    if(flag)    cout<<"Yes\n";
    else        cout<<"No\n";
    return 0;
}