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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n, i, m, x;
    cin >> n >> m;
    long long a[n];
    for (auto &ghe : a) cin >> ghe;
    ;
    priority_queue<long long> pq[200001];
    for (auto &i : a) {
      x = 0;
      while (i) pq[i].push(-x), i >>= 1, ++x;
      pq[0].push(-x);
    }
    x = LLONG_MAX;
    for (auto &q : pq)
      if (q.size() >= m) {
        i = m;
        n = 0;
        while (i--) n -= q.top(), q.pop();
        x = min(x, n);
      }
    cout << x << endl;
  }
  return 0;
}