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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> pod(2 * 1e5 + 2);
  vector<vector<long long>> all(2 * 1e5 + 2);
  for (long long i = 0; i < n; i++) {
    long long x = a[i];
    long long cnt = 0;
    while (x >= 0) {
      pod[x]++;
      all[x].push_back(cnt);
      if (x == x / 2) {
        break;
      }
      x /= 2;
      cnt++;
    }
  }
  vector<long long> anses;
  for (long long i = pod.size() - 1; i >= 0; i--) {
    if (pod[i] >= k) {
      anses.push_back(i);
    }
  }
  long long result = 1e9;
  for (long long j = 0; j < anses.size(); j++) {
    long long ans = anses[j];
    sort(all[ans].begin(), all[ans].end());
    long long res = 0;
    for (long long i = 0; i < k; i++) {
      res += all[ans][i];
    }
    result = min(result, res);
  }
  cout << result;
  return 0;
}