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>
#pragma GCC optimize(2)
using namespace std;
const long long mod = 998244353;
const long double eps = 1e-9;
const long double PI = acos(-1.0);
long long n, k, ans = 223372036854775807, num[501000], s[501000];
map<long long, long long> mp;
int main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> num[i];
  for (int i = 1; i <= n; i++) {
    mp[num[i]]++;
    if (mp[num[i]] >= k) {
      cout << 0;
      return 0;
    }
  }
  sort(num + 1, num + 1 + n);
  for (long long i = 1; i <= n; i++) s[i] = s[i - 1] + num[i];
  for (long long i = 1; i <= n; i++) {
    long long buf;
    if (num[i] != num[i - 1]) {
      buf = s[n] - s[i] - (n - i) * num[i];
      if (n - i + 1 < k)
        buf += (i - 1) * (num[i] - 1) - s[i - 1] + k - (n - i + 1);
      ans = min(ans, buf);
      if (i - 1 + mp[num[i]] >= k)
        buf = (i - 1) * (num[i] - 1) - s[i - 1] + (k - mp[num[i]]);
      ans = min(ans, buf);
    }
    if (num[i] != num[i + 1]) {
      long long buf = (i - 1) * num[i] - s[i - 1];
      if (i < k) buf += s[n] - s[i] - (n - i) * (num[i] + 1) + k - i;
      ans = min(ans, buf);
      if (n - i + mp[num[i]] >= k)
        buf = s[n] - s[i] - (n - i) * (num[i] + 1) + (k - mp[num[i]]);
      ans = min(ans, buf);
    }
  }
  cout << ans;
  return 0;
}