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 n;
  int k;
  cin >> n >> k;
  int ans[200009];
  int a[n];
  memset(ans, -1, sizeof(ans));
  int max1 = INT_MIN;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    max1 = max(max1, a[i]);
  }
  long long int count = 0;
  long long int count1 = 0;
  int q = 0;
  sort(a, a + n);
  int flag = 0;
  for (int i = 1; i <= max1; i++) {
    count1 = 0;
    q = 0;
    for (int j = 0; j < n; j++) {
      int x = a[j];
      count = 0;
      flag = 0;
      while (x > i) {
        x = x / 2;
        count++;
      }
      if (x == i) {
        count1 += count;
        q++;
        if (q == k) {
          flag = 1;
          break;
        }
      }
    }
    if (flag == 1) ans[i] = count1;
  }
  int r = INT_MAX;
  for (int i = 1; i <= max1; i++) {
    if (ans[i] != -1) r = min(r, ans[i]);
  }
  cout << r << " ";
}