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(NULL);
  cout.tie(NULL);
  ;
  long long int n, k, i;
  cin >> n >> k;
  long long int a[n];
  map<long long int, long long int> m;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  sort(a, a + n);
  long long int ans[200005] = {};
  for (i = 0; i < n; i++) {
    if (k <= m[a[i]]) {
      cout << 0;
      return 0;
    }
    long long int x = a[i], y = 0;
    while (x) {
      x /= 2;
      y++;
      if (m.find(x) == m.end() || m[x] < k) {
        m[x]++;
        ans[x] += y;
      }
    }
  }
  long long int mins = LLONG_MAX;
  for (i = 0; i < 200005; i++) {
    if (m.find(i) != m.end() && m[i] >= k) mins = min(mins, ans[i]);
  }
  cout << mins;
}