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)

#include <bits/stdc++.h>
using namespace std;
#define ll long long int 
#define ull unsigned long long int
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define mx *max_element
#define mn *min_element
#define fo(i,n) for(ll i=0;i<n;i++)
#define of(i,p,n) for(ll i=n-1;i>=p;i--)
#define all(x) x.begin(),x.end()
#define sortall(x) sort(all(x))
#define mem(x,p) memset(x,p,sizeof(x))
#define trav(it,a) for(auto it=a.begin();it!=a.end();++it)
#define kk "\n"
const ll mod=1e9+7;
int main()
{
  IOS
  ll n,m;
  cin>>n>>m;
  ll arr[n],s=0;
  fo(i,n)
  {
    cin>>arr[i];
    s=s+arr[i];
  }
  sort(arr,arr+n);
  ll p=s/(4*m);ll c=0;
  fo(i,n)
  {
    if(arr[i]>=p)
      c++;
  }
  if(c>=m)
    cout<<"Yes";
  else
    cout<<"No";
  return 0;
}