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 long long mod = 1e9 + 7;
const long long N = 2e5 + 5;
const long long inf = 0x3f3f3f3f;
int a[N];
int num[N];
int op[N];
int main() {
  std::ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int ans = inf;
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int t = a[i];
    num[t]++;
    if (num[t] >= k) ans = min(op[t], ans);
    int g = 1;
    while (t != 0) {
      t /= 2;
      num[t]++;
      op[t] += g;
      g++;
      if (num[t] >= k) ans = min(op[t], ans);
    }
  }
  cout << ans << endl;
  return 0;
}