Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long arr[200005], sh[200005], sp[200005];
int main() {
  long long i, j, k, n, m, p, q, t, cnt = 0, mn;
  scanf("%lld", &n);
  scanf("%lld", &k);
  for (i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
    ;
  }
  sort(arr, arr + n);
  for (i = 0; i < n; i++) {
    cnt = 0;
    while (1) {
      sh[arr[i]]++;
      if (sh[arr[i]] <= k) sp[arr[i]] += cnt;
      if (!arr[i]) break;
      arr[i] /= 2;
      cnt++;
    }
  }
  mn = sp[0];
  for (i = 0; i <= 200005; i++) {
    if (sh[i] >= k && sp[i] < mn) mn = sp[i];
  }
  cout << mn << endl;
  return 0;
}