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;
const int N = 200000;
long long a[N + 5], sum[N + 5];
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) sum[i] = a[i] + sum[i - 1];
  long long ans = LLONG_MAX;
  for (int i = 1, j; i <= n; i = j) {
    for (j = i; j <= n; ++j) {
      if (a[j] != a[i]) break;
    }
    long long p = (i - 1) * (a[i] - 1) - sum[i - 1];
    long long q = sum[n] - sum[j - 1] - (n - j + 1) * (a[i] + 1);
    vector<pair<long long, int>> vc;
    vc.push_back({p, i - 1});
    vc.push_back({q, (n - j + 1)});
    vc.push_back({p + q, n - (j - i)});
    sort(vc.begin(), vc.end());
    int t = k - (j - i);
    long long c = 0;
    if (t > 0)
      for (auto &it : vc) {
        if (t <= it.second) {
          c = it.first + t;
          break;
        }
      }
    ans = min(ans, c);
  }
  printf("%lld\n", ans);
  return 0;
}
