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;
int main() {
  int t = 1;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> v(n);
    for (int i = 0; i < n; ++i) cin >> v[i];
    map<int, int> mp;
    map<int, int> step;
    sort(v.begin(), v.end());
    for (int i = 0; i < n; ++i) {
      int x = v[i];
      int a = 0;
      while (x > 0) {
        if (mp[x] < k) {
          mp[x]++;
          step[x] += a;
        }
        x /= 2;
        ++a;
      }
      mp[0]++;
      step[0] += a;
    }
    int ans = 2e9;
    for (int i = 0; i <= 200000; ++i) {
      if (mp[i] >= k) ans = min(ans, step[i]);
    }
    cout << ans << "\n";
  }
}