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 a[202020];
long long l[202020], r[202020];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 2; i <= n; i++)
    if (a[i] == a[i - 1])
      l[i] = l[i - 1];
    else
      l[i] = l[i - 1] + 1LL * (i - 1) * (a[i] - a[i - 1]);
  for (int i = n - 1; i >= 1; i--)
    if (a[i] == a[i + 1])
      r[i] = r[i + 1];
    else
      r[i] = r[i + 1] + 1LL * (n - i) * (a[i + 1] - a[i]);
  long long ans = (1LL << 62);
  for (int i = 1; i <= n; i++) {
    int save = i;
    while (a[i] == a[i + 1]) i++;
    int cnt = i - save + 1;
    if (cnt >= k) {
      ans = 0;
      break;
    }
    long long ans1 = r[i + 1] + 1LL * (a[i + 1] - a[i] - 1) * (n - i);
    int cnt1 = n - i;
    if (cnt + cnt1 >= k) {
      ans1 += k - cnt;
    } else {
      ans1 += l[save - 1] + 1LL * (a[save] - a[save - 1] - 1) * (save - 1);
      ans1 += k - cnt;
    }
    long long ans2 =
        l[save - 1] + 1LL * (a[save] - a[save - 1] - 1) * (save - 1);
    int cnt2 = save - 1;
    if (cnt + cnt2 >= k)
      ans2 += k - cnt;
    else {
      ans2 += r[i + 1] + 1LL * (a[i + 1] - a[i] - 1) * (n - i);
      ans2 += k - cnt;
    }
    ans = min(ans, min(ans1, ans2));
  }
  printf("%lld\n", ans);
  return 0;
}