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 long long N = 2e5 + 7;
const long long M = 1e9 + 7;
const long long Mod = 998244353;
long long _, i, j, k, n, m, t, T, p, q, len, u, v, w, st, en, x, y, ma, mi, num,
    flag, res, cnt, ans, tmp1, tmp2, nex;
long long a[N], s1[N], s2[N];
map<long long, long long> l, r;
signed main() {
  ios::sync_with_stdio(false);
  long long t = 1;
  while (t--) {
    cin >> n >> k;
    for (i = 1; i <= n; i++) cin >> a[i];
    sort(a + 1, a + n + 1);
    for (i = 1; i <= n; i++)
      if (!l[a[i]]) l[a[i]] = i;
    for (i = n; i >= 1; i--)
      if (!r[a[i]]) r[a[i]] = i;
    for (i = 1; i <= n; i++) s1[i] = s1[i - 1] + a[i];
    for (i = n; i >= 1; i--) s2[i] = s2[i + 1] + a[i];
    ans = 1e18;
    for (i = 1; i <= n; i++) {
      if (r[a[i]] - l[a[i]] + 1 >= k) {
        ans = 0;
        break;
      }
      tmp1 = (a[i] - 1) * (l[a[i]] - 1) - s1[l[a[i]] - 1];
      tmp2 = s2[r[a[i]] + 1] - (a[i] + 1) * (n - r[a[i]]);
      if (i >= k) ans = min(ans, tmp1 + k - r[a[i]] + l[a[i]] - 1);
      if (n - i + 1 >= k) ans = min(ans, tmp2 + k - (r[a[i]] - l[a[i]] + 1));
      ans = min(ans, tmp1 + tmp2 + k - (r[a[i]] - l[a[i]] + 1));
    }
    cout << ans;
  }
  return 0;
}