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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " , ";
  __f(comma + 1, args...);
}
const int INF = 1e16;
const int nax = 2e5 + 5;
vector<int> v[nax];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    int cnt = 0;
    while (x) {
      v[x].push_back(cnt);
      x /= 2;
      cnt++;
    }
  }
  int ans = INF;
  for (int i = 1; i < nax; i++) {
    sort(v[i].begin(), v[i].end());
    int sum = INF;
    if (v[i].size() >= k) {
      sum = 0;
      for (int j = 0; j < k; j++) {
        sum += v[i][j];
      }
    }
    ans = min(ans, sum);
  }
  cout << ans << endl;
  return 0;
}