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;
int n, k, res = 1e9, ma;
int d[3][500005], a[500005];
vector<int> b[100005];
map<int, int> s[200005];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    ma = max(ma, a[i]);
    d[1][a[i]]++;
    int t = a[i];
    int f = 0;
    while (t != 0) {
      f++;
      d[2][t / 2]++;
      s[t / 2][d[2][t / 2]] = s[t / 2][d[2][t / 2] - 1] + f;
      t = t / 2;
    }
  }
  for (int i = 0; i <= ma; i++) {
    if (d[1][i] + d[2][i] >= k) {
      if (d[1][i] >= k)
        res = min(res, 0);
      else
        res = min(res, s[i][k - d[1][i]]);
    }
  }
  cout << res;
}