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() {
  long long n, k;
  cin >> n >> k;
  long long a[n + 10];
  set<int> vect;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    long long x = a[i];
    while (x > 0) {
      vect.insert(x);
      x /= 2;
    }
  }
  long long ans = 1e10;
  set<int>::iterator it;
  for (it = vect.begin(); it != vect.end(); it++) {
    vector<int> cnt;
    for (long long j = 0; j < n; j++) {
      long long op = 0;
      long long x = a[j];
      while (x > (*it)) {
        x = x / 2;
        op += 1;
      }
      if (x == (*it)) cnt.push_back(op);
    }
    if (cnt.size() < k) {
      continue;
    }
    sort(cnt.begin(), cnt.end());
    ans = min(ans, (long long)accumulate(cnt.begin(), cnt.begin() + k, 0));
  }
  cout << ans << "\n";
  return 0;
}