You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and the required number of equal elements.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4
#include <bits/stdc++.h>
using namespace std;
long long a[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, k;
  while (cin >> n >> k) {
    map<long long, long long> mp, mp1;
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      mp[a[i]]++;
      if (mp[a[i]] >= k) {
        ans = 1;
      }
    }
    if (ans) {
      cout << "0" << endl;
      continue;
    }
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= k; i++) {
      ans += (a[i] / 2) + 1;
    }
    for (int i = 1; i <= n; i++) {
      long long num = a[i], x = 0;
      while (num > 0) {
        num = num / 2;
        x++;
        mp[num]++;
        mp1[num] += x;
        if (mp[num] >= k) {
          ans = min(ans, mp1[num]);
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
