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 mxn = 2e5 + 10;
int a[mxn];
int main() {
  map<int, int> mark, ans;
  map<int, int>::iterator it;
  int n, k;
  cin >> n >> k;
  int a[n + 2];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mark[a[i]]++;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++)
    if (mark[a[i]] >= k) {
      cout << 0 << endl;
      return 0;
    }
  for (int i = 0; i < n; i++) {
    int op = 0;
    while (a[i] > 0) {
      a[i] /= 2;
      op++;
      if (mark[a[i]] < k) {
        mark[a[i]]++;
        ans[a[i]] += op;
      }
    }
  }
  int res = 1e9;
  for (it = ans.begin(); it != ans.end(); it++)
    if (mark[it->first] == k) res = min(res, it->second);
  cout << res << endl;
}