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 n, k, ans = 1e9;
vector<int> a(200000 + 10);
vector<vector<int> > v(200000 + 10);
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    int x = a[i], cur = 0;
    while (x > 0) {
      v[x].push_back(cur);
      x /= 2;
      cur++;
    }
  }
  for (int i = 0; i <= 200000; i++) {
    sort(v[i].begin(), v[i].end());
    if (int(v[i].size()) < k) continue;
    ans = min(ans, accumulate(v[i].begin(), v[i].begin() + k, 0));
  }
  printf("%d\n", ans);
  return 0;
}