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<int> A[2 * 100005];
int main() {
  int n, s;
  cin >> n >> s;
  int arr[2 * 100005];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  for (int i = 0; i < n; i++) {
    int t = arr[i];
    int step = 0;
    A[t].push_back(step++);
    while (t) {
      t >>= 1;
      A[t].push_back(step++);
    }
  }
  int Min = 9999999;
  for (int i = 0; i <= 2 * 100000; i++) {
    if (A[i].size() < s) continue;
    int ans = 0;
    for (int j = 0; j < s; j++) {
      ans += A[i][j];
      if (ans > Min) break;
    }
    if (Min > ans) Min = ans;
  }
  cout << Min << endl;
  return 0;
}