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 int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) res = res * x;
    x = x * x;
    n = n / 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k, i, j;
  cin >> n >> k;
  int a[n];
  int f[2 * 100002] = {0};
  int ans[2 * 100002] = {0};
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    int x = a[i], cnt = 0;
    while (x > 0) {
      if (f[x] < k) {
        f[x]++;
        ans[x] += cnt;
      }
      x = x / 2;
      cnt++;
    }
    if (f[0] < k) {
      f[0]++;
      ans[0] += cnt;
    }
  }
  int y = INT_MAX;
  for (i = 0; i <= 2 * 100001; i++) {
    if (f[i] == k) {
      y = min(ans[i], y);
    }
  }
  cout << y;
}