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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
int a[200005] = {0}, ci[200005] = {0}, now[200005] = {0};
int main() {
  int n, kk;
  scanf("%d", &n), scanf("%d", &kk);
  int ans = 0x3f3f3f3f;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    now[a[i]]++;
    if (now[a[i]] >= kk) ans = 0;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int temp = a[i];
    int k = 1;
    while (temp != 0) {
      temp /= 2;
      ci[temp] += k;
      now[temp]++;
      if (now[temp] >= kk) ans = min(ci[temp], ans);
      k++;
    }
  }
  printf("%d", ans);
}
