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);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  map<long long, long long> freq;
  map<long long, long long> m;
  vector<long long> v(n, 0);
  long long ans = 1000000007;
  for (int i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    freq[v[i]]++;
    if (freq[v[i]] >= k) ans = min(ans, m[v[i]]);
    long long x = v[i];
    if (x == 0) continue;
    long long cnt = 0;
    while (1) {
      x = x / 2;
      freq[x]++;
      cnt++;
      m[x] += cnt;
      if (freq[x] >= k) ans = min(ans, m[x]);
      if (x == 0) break;
    }
  }
  cout << ans;
  return 0;
}