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;
int ara[51];
map<int, int> mp, mo;
int main() {
  int n, i, mn, c, ans, k;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", &ara[i]);
    mp[ara[i]]++;
  }
  sort(ara, ara + n);
  mn = 0;
  ans = -1;
  for (i = 0; i < n; i++) {
    c = 0;
    if (mp[ara[i]] >= k) {
      if (ans == -1)
        ans = mo[ara[i]];
      else
        ans = min(ans, mo[ara[i]]);
    }
    while (true) {
      c++;
      ara[i] = ara[i] >> 1;
      mp[ara[i]]++;
      mo[ara[i]] += c;
      if (mp[ara[i]] >= k) {
        if (ans == -1)
          ans = mo[ara[i]];
        else
          ans = min(ans, mo[ara[i]]);
      }
      if (ara[i] == 0) break;
    }
  }
  printf("%d\n", ans);
  return 0;
}