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 dx[] = {0, -1, 0, 1, -1, 1, -1, 1};
int dy[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int MOD = 1e9 + 7, N = 2e5 + 5, oo = 1e9;
const double pi = acos(-1);
using namespace std;
long long freq[N], f[N][100];
void solve(long long n) {
  long long cnt = 0;
  while (n >= 1) {
    f[n][cnt]++;
    n /= 2;
    cnt++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int k, n;
  cin >> n >> k;
  long long arr[n + 5];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    freq[arr[i]]++;
    solve(arr[i]);
  }
  sort(arr, arr + n);
  long long final = 1e18;
  for (int i = 1; i < N; i++) {
    long long add = 0;
    long long ans = 0;
    for (int j = 0; j < 50; j++) {
      ans += j * min(f[i][j], k - add);
      add += f[i][j];
      if (add >= k) {
        final = min(final, ans);
        break;
      }
    }
  }
  cout << final;
  return 0;
}