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
#include <bits/stdc++.h>
using namespace std;
multiset<long long> s[200004];
long long a[200005];
long long su[200005];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long x;
  for (long long i = 0; i < n; i++) {
    x = 0;
    while (true) {
      s[a[i]].insert(x);
      su[a[i]] += x;
      if (s[a[i]].size() > k) {
        set<long long>::iterator pointer = s[a[i]].end();
        pointer--;
        su[a[i]] -= *pointer;
        s[a[i]].erase(pointer);
      }
      if (a[i] == 0) break;
      a[i] /= 2;
      x++;
    }
  }
  long long minn = 1e9;
  for (long long i = 0; i < 200005; i++) {
    if (s[i].size() == k) {
      minn = min(minn, su[i]);
    }
  }
  cout << minn;
  return 0;
}
