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() {
  long long int n, k, i;
  cin >> n >> k;
  long long int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  map<long long int, vector<long long int> > m;
  for (i = 0; i < n; i++) {
    long long int c = a[i], cnt = 0;
    while (c > 0) {
      m[c].push_back(cnt);
      c = c / 2;
      cnt++;
    }
  }
  long long int ans = 0, mini = 100000000000;
  for (auto it = m.begin(); it != m.end(); it++) {
    ans = 0;
    if (it->second.size() >= k) {
      for (i = 0; i < k; i++) ans = ans + it->second[i];
      mini = min(ans, mini);
    }
  }
  cout << mini;
}