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;
const int N = 2e5 + 10;
long long a[N], pre[N], suf[N], cnt[N];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
  for (int i = 1; i <= n; i++) pre[i] = i * a[i] - pre[i];
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + a[i];
  for (int i = n; i >= 1; i--) suf[i] = suf[i] - a[i] * (n - i + 1);
  for (int i = 1; i <= n; i++) cnt[i] = (a[i] == a[i - 1] ? cnt[i - 1] + 1 : 1);
  long long ans = 0x3f3f3f3f3f3f3f3fll;
  for (int i = 1; i <= n; i++)
    if (cnt[i] >= k) {
      printf("0\n");
      return 0;
    }
  for (int i = 1; i <= n; i++) {
    if (i < k)
      ans = min(ans, pre[i] + suf[i] - (n - k));
    else
      ans = min(ans, pre[i] - (i - k));
    if (i > n - k + 1)
      ans = min(ans, suf[i] + pre[i] - (n - k));
    else
      ans = min(ans, suf[i] - (n - i + 1 - k));
  }
  printf("%lld\n", ans);
  return 0;
}