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>
int n, k;
int a[100001];
int b[100001];
int min = 0x3f3f3f3f;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    b[a[i]]++;
  }
  std::sort(a + 1, a + n + 1);
  for (int i = 0; i <= 200001; ++i) {
    int ans = i, answer = 0, need = k;
    for (int j = 1; j <= n; ++j) {
      int kk = a[j];
      int step = 0;
      while (kk > ans) kk /= 2, ++step;
      if (kk == ans && need) answer += step, --need;
      if (!need) break;
    }
    if (!need) min = std::min(answer, min);
  }
  printf("%d", min);
}