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 cnt[200010];
int num[200010];
int a[60];
int main() {
  int n, k;
  while (~scanf("%d%d", &n, &k)) {
    memset(a, 0, sizeof(a));
    memset(num, 0, sizeof(num));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    sort(a, a + n);
    int ans = 0x3f3f3f3f;
    for (int i = 0; i < n; i++) {
      int x = a[i];
      int d = 0;
      while (x) {
        num[x]++;
        cnt[x] += d;
        if (num[x] == k) {
          ans = min(ans, cnt[x]);
        }
        x /= 2;
        d++;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}