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 maxn = 2e5 + 5;
long long a[maxn];
long long sum[maxn];
inline long long getsum(int l, int r) {
  if (l > r)
    return 0ll;
  else
    return sum[r] - sum[l - 1];
}
int main(void) {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
  long long res = 1e18;
  for (int i = 1; i <= n; i++) {
    int j = i;
    while (j + 1 <= n && a[j + 1] == a[i]) j++;
    int d1 = i - 1;
    int d2 = j - i + 1;
    int d3 = n - j;
    long long s1, s2, t;
    if (d2 >= k) {
      res = 0;
      break;
    }
    if (d1 + d2 >= k) {
      s1 = getsum(1, i - 1);
      t = 1ll * d1 * (a[i] - 1) - s1;
      res = min(res, t + (k - d2));
    }
    if (d3 + d2 >= k) {
      s2 = getsum(j + 1, n);
      t = s2 - 1ll * d3 * (a[i] + 1);
      res = min(res, t + (k - d2));
    }
    s1 = getsum(1, i - 1);
    long long t1 = 1ll * d1 * (a[i] - 1) - s1;
    s2 = getsum(j + 1, n);
    long long t2 = s2 - 1ll * d3 * (a[i] + 1);
    res = min(res, t1 + t2 + (k - d2));
    i = j;
  }
  printf("%I64d\n", res);
  return 0;
}
