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;
long long cnt[3000000];
long long cost[3000000];
int main() {
  long long n, k;
  cin >> n >> k;
  long long ans = 1e9;
  vector<long long> inputs;
  for (long long i = 0; i < n; ++i) {
    long long a;
    cin >> a;
    inputs.push_back(a);
  }
  sort(inputs.begin(), inputs.end());
  for (long long i = 0; i < n; ++i) {
    long long a = inputs[i];
    long long now_cost = 0;
    while (a != 0) {
      cnt[a]++;
      cost[a] += now_cost;
      if (cnt[a] >= k) {
        ans = min(ans, cost[a]);
      }
      now_cost++;
      a /= 2;
    }
  }
  cout << ans << endl;
}