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;
long long read() {
  long long num = 0;
  bool f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    f = (ch == '-');
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num = (num << 1) + (num << 3) + ch - '0';
    ch = getchar();
  }
  return f ? -num : num;
}
long long n, k, t, ans;
long long a[222222], sum[222222], c[222222];
const long long INF = (1ll << 62) - 1ll;
int main() {
  n = read(), k = read();
  ans = INF;
  t = 1ll;
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + a[i];
    c[i] = (a[i] == a[i - 1]) ? c[i - 1] + 1 : 1;
    if (c[i] == k) {
      puts("0");
      return 0;
    }
  }
  while (t <= n) {
    long long tmp1 = INF, tmp2 = INF, tmp3 = INF;
    if (t >= k) tmp1 = a[t] * t - sum[t] - (t - k);
    if (n - t + 1 >= k)
      tmp2 = sum[n] - sum[t - 1] - a[t] * (n - t + 1) - (n - t + 1 - k);
    tmp3 =
        a[t] * t - sum[t] + sum[n] - sum[t - 1] - a[t] * (n - t + 1) - (n - k);
    ans = min(ans, min(tmp1, min(tmp2, tmp3)));
    t++;
  }
  printf("%lld\n", ans);
  return 0;
}
