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;
long long expo(long long x, long long n, long long m) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return expo((x * x) % m, n / 2, m);
  else
    return (x * expo((x * x) % m, (n - 1) / 2, m)) % m;
}
int arr[200005];
int steps[200005];
void solve() {
  vector<int> v;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    v.push_back(t);
  }
  int a;
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    a = v[i];
    arr[a] = arr[a] + 1;
    int j = 1;
    while (a != 0) {
      a /= 2;
      arr[a]++;
      if (arr[a] <= k) {
        steps[a] += j;
      }
      j++;
    }
  }
  int ans = 2147483647;
  for (int i = 1; i <= 200005; i++)
    if (arr[i] >= k) ans = min(ans, steps[i]);
  cout << ans << endl;
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}