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 maxx = 2e5 + 6;
long long a[maxx];
long long pre[maxx];
long long bac[maxx];
int main() {
  int k, n;
  while (~scanf("%d%d", &n, &k)) {
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    sort(a + 1, a + 1 + n);
    pre[0] = 0;
    for (int i = 1; i <= n; i++) {
      pre[i] = pre[i - 1] + a[i];
    }
    bac[n + 1] = 0;
    for (int i = n; i >= 1; i--) {
      bac[i] = bac[i + 1] + a[i];
    }
    long long ans = 1e18;
    int max_cnt = 0;
    int cnt = 1;
    for (int i = 1; i <= n; i++) {
      if (i != 1 && a[i] == a[i - 1]) {
        cnt++;
      } else {
        cnt = 1;
      }
      max_cnt = max(max_cnt, cnt);
    }
    if (max_cnt >= k) {
      printf("0\n");
      continue;
    }
    for (long long i = 1; i <= n; i++) {
      long long l = i * a[i] - pre[i];
      long long r = bac[i] - (n - i + 1) * a[i];
      if (i >= k) {
        ans = min(ans, l - (i - k));
      }
      if (n - i + 1 >= k) {
        ans = min(ans, r - (n - i + 1 - k));
      }
      ans = min(ans, l + r - (n - k));
    }
    printf("%lld\n", ans);
  }
  return 0;
}