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
#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 5;
int n, k, a;
int vis_time[MAX], vis_cnt[MAX], ans;
struct data {
  int size, op;
  bool operator<(data b) const {
    if (size == b.size)
      return op > b.op;
    else
      return size < b.size;
  }
};
priority_queue<data> line;
int main() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; ++i) {
    scanf("%d", &a);
    line.push((data){a, 0});
  }
  ans = 0x3f3f3f3f;
  while (!line.empty()) {
    data tmp = line.top();
    line.pop();
    int size = tmp.size, op = tmp.op;
    vis_time[size]++;
    vis_cnt[size] += op;
    if (vis_time[size] == k) {
      ans = min(ans, vis_cnt[size]);
    }
    if (!size)
      continue;
    else
      line.push((data){size / 2, op + 1});
  }
  printf("%d\n", ans);
  return 0;
}
