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, m, ans = INT_MAX;
int steps[200005] = {0}, freq[200005] = {0};
void cal(int x, int step) {
  freq[x]++;
  if (freq[x] == k) ans = min(ans, steps[x]);
  if (x == 0) return;
  step++;
  steps[x / 2] += step;
  cal(x / 2, step);
}
int main() {
  cin >> n >> k;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr.begin(), arr.end());
  for (int i = 0; i < n; i++) cal(arr[i], 0);
  cout << ans << endl;
  return 0;
}