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>
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
int n;
int k;
int tot[N];
int a[N];
int ans[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    tot[a[i]]++;
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; ++i) {
    int tmp = a[i];
    int cur = 0;
    bool flag = false;
    while (tmp >= 0) {
      if (flag) break;
      tmp /= 2;
      cur++;
      if (tot[tmp] < k) {
        tot[tmp]++;
        ans[tmp] += cur;
      }
      if (tmp == 0) flag = true;
    }
  }
  int res = INF;
  for (int i = 0; i <= 200010; ++i) {
    if (tot[i] >= k) res = min(res, ans[i]);
  }
  cout << res << '\n';
  return 0;
}