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>
const double EPS = 0.00000001;
const long long mod = 1000000000 + 7;
using namespace std;
map<int, int> a[200020];
int main() {
  ios::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[x][0]++;
  }
  long long N = 200000, ans = 10000000000LL;
  for (int i = N; i >= 1; i--) {
    if (i == 5) {
    }
    long long cnt = 0, cost = 0;
    for (auto p : a[i]) {
      int price = p.first, num = p.second;
      a[i / 2][price + 1] += num;
      if (num + cnt >= k) {
        cost += (k - cnt) * price;
        cnt = k;
        break;
      } else {
        cost += price * num;
        cnt += num;
      }
    }
    if (cnt >= k) ans = min(ans, cost);
  }
  cout << ans << endl;
  return 0;
}