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 int N = 2e5;
vector<long long> adj[N + 3];
int main() {
  long long n, k, ans = 1e12;
  cin >> n >> k;
  long long a[n + 2];
  vector<long long> valu;
  valu.push_back(0);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    long long val = a[i];
    long long cnt = 0;
    while (val >= 0) {
      adj[val].push_back(cnt);
      if (val == 0) break;
      cnt++;
      val /= 2;
    }
  }
  for (long long i = 0; i <= N; i++) {
    sort(adj[i].begin(), adj[i].end());
    if (adj[i].size() < k) continue;
    long long sum = 0;
    for (long long j = 0; j < k; j++) {
      sum += adj[i][j];
    }
    ans = min(sum, ans);
  }
  cout << ans << endl;
}