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 int check(long long int n, long long int m) {
  long long int ans = 0;
  while (n > m) {
    n = n / 2;
    ans++;
  }
  if (n == m) {
    return ans;
  }
  return -1;
}
int32_t main() {
  long long int n, k;
  scanf("%lld", &n);
  scanf("%lld", &k);
  vector<long long int> v(n);
  long long int maxi = -1;
  for (long long int i = 0; i < n; ++i) {
    scanf("%lld", &v[i]);
    maxi = max(maxi, v[i]);
  }
  long long int ans = 1e18;
  sort(v.begin(), v.end());
  for (long long int i = 0; i <= maxi; ++i) {
    long long int ans1 = 0, k1 = 0;
    for (long long int j = 0; j < n; ++j) {
      long long int x = check(v[j], i);
      if (x != -1) {
        ans1 += x;
        k1++;
        if (k1 == k) {
          ans = min(ans, ans1);
          break;
        }
      }
    }
  }
  printf("%lld", ans);
  return 0;
}
