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;
vector<pair<int, int> > cost[200020];
int a[200020];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    cost[a[i]].push_back(pair<int, int>(a[i], 0));
    if (cost[a[i]].size() >= k) {
      puts("0");
      return 0;
    }
  }
  int ans = 987654321, cnt = 1;
  vector<int> ansN;
  bool success = false;
  while (cnt <= 20) {
    for (int i = 0; i < n; i++) {
      if (a[i] != 0) {
        a[i] /= 2;
        cost[a[i]].push_back(pair<int, int>(a[i], cnt));
      }
      if (cost[a[i]].size() >= k) {
        success = true;
        ansN.push_back(a[i]);
      }
    }
    if (success) {
      for (int j = 0; j < ansN.size(); j++) {
        int tmp = 0;
        for (int i = 0; i < k; i++) {
          tmp += cost[ansN[j]][i].second;
        }
        ans = min(ans, tmp);
      }
    }
    cnt++;
  }
  printf("%d\n", ans);
}