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;
int n, k, a[N];
long long s[N], ans = 1e18, pre[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (int i = n; i >= 1; i--) pre[i] = pre[i + 1] + a[i];
  for (int i = 1; i <= n; i++) {
    s[i] = s[i - 1] + a[i];
    if (i >= k && a[i] == a[i - k + 1]) return cout << 0, 0;
  }
  for (int i = 1, j; i <= n; i = j) {
    j = i;
    while (j + 1 <= n && a[j + 1] == a[i]) ++j;
    int x = i - 1, y = n - j, z = j - i + 1;
    if (z + x >= k) {
      ans = min(ans, -s[x] + 1ll * x * a[i] - (x - (k - z)));
    }
    if (y + z >= k) {
      ans = min(ans, pre[j + 1] - 1ll * y * a[j] - (y - (k - z)));
    }
    ans = min(ans, -s[x] + 1ll * x * a[i] + pre[j + 1] - 1ll * y * a[j] -
                       (x + y - (k - z)));
    ++j;
  }
  cout << ans << '\n';
  return 0;
}