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 a[200005];
vector<int> mg[200005];
vector<int> b;
int main() {
  int n, k, i, sum, res = 100000000, j, t, x;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    x = a[i];
    t = 1;
    mg[x].push_back(0);
    while (x > 0) {
      x = x / 2;
      mg[x].push_back(t);
      t++;
    }
  }
  for (i = 0; i < 200000; i++) {
    if (mg[i].size() < k) continue;
    int kol = mg[i].size();
    sort(mg[i].begin(), mg[i].end());
    sum = 0;
    for (j = 0; j < k; j++) sum += mg[i][j];
    res = min(res, sum);
  }
  cout << res << endl;
  return 0;
}