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;
int32_t main() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  long long i, j;
  long long value = -10;
  for (i = 0; i < n; i++) cin >> a[i];
  long long counting[200001];
  long long totalOperations[200001];
  for (i = 0; i < 200001; i++) totalOperations[i] = 0;
  long long count = 0;
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    if (a[i] == 0) {
      counting[0]++;
    } else {
      counting[a[i]]++;
      count = 0;
      while (a[i] != 0) {
        a[i] = a[i] / 2;
        count++;
        if (counting[a[i]] < k)
          totalOperations[a[i]] = totalOperations[a[i]] + count;
        counting[a[i]]++;
      }
    }
  }
  long long oper = INT_MAX;
  for (i = 0; i < 200001; i++) {
    if (counting[i] >= k) {
      if (oper > totalOperations[i]) oper = totalOperations[i];
    }
  }
  cout << oper << endl;
}
