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(0);
  cin.tie(0);
  int n, k;
  vector<int> A[200001];
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int x, cnt = 0;
    cin >> x;
    while (x) {
      A[x].push_back(cnt++);
      if (x == 0) break;
      x >>= 1;
    }
  }
  int ans = 1e9;
  for (int i = 0; i <= 200000; ++i) {
    sort(A[i].begin(), A[i].end());
    if (A[i].size() < k) continue;
    int sum = 0;
    for (int j = 0; j < k; ++j) sum += A[i][j];
    ans = min(ans, sum);
  }
  printf("%d\n", ans);
  return 0;
}