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 maxn = 2e5 + 5;
long long a[maxn];
long long s[maxn], ans, n, k;
bool cmp(int x, int y) { return x < y; }
int main() {
  long long i, j, d;
  long long ans1, ans2;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1, cmp);
  for (i = 1; i <= n; i++) s[i] = s[i - 1] + a[i];
  ans = s[n];
  for (i = 1; i <= n; i = j + 1) {
    for (j = i; j <= n; j++)
      if (a[j] > a[i]) break;
    j--;
    if (j - i + 1 >= k) {
      printf("0\n");
      return 0;
    }
    d = k - (j - i + 1);
    if (i - 1 >= d && n - j >= d) {
      ans1 = a[i] * (i - 1) - s[i - 1] - (i - 1);
      ans2 = s[n] - s[j] - (n - j) * a[j] - (n - j);
      ans = min(ans, min(ans1, ans2) + d);
    }
    if (i - 1 >= d && n - j < d) {
      ans1 = a[i] * (i - 1) - s[i - 1] - (i - 1);
      ans = min(ans, ans1 + d);
    }
    if (i - 1 < d && n - j >= d) {
      ans2 = s[n] - s[j] - (n - j) * a[j] - (n - j);
      ans = min(ans, ans2 + d);
    }
    if (i - 1 < d && n - j < d) {
      ans1 = a[i] * (i - 1) - s[i - 1] - (i - 1);
      ans2 = s[n] - s[j] - (n - j) * a[j] - (n - j);
      ans = min(ans, ans1 + ans2 + d);
    }
  }
  printf("%lld\n", ans);
  return 0;
}