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 main() {
  std::size_t n, k;
  std::cin >> n >> k;
  unsigned constexpr MAX = 2e5 + 5;
  std::vector<std::vector<unsigned>> times(MAX);
  while (n--) {
    unsigned x;
    std::cin >> x;
    unsigned cur = 0;
    while (true) {
      times[x].push_back(cur);
      if (!x) {
        break;
      }
      x /= 2;
      ++cur;
    }
  }
  unsigned result = std::numeric_limits<unsigned>::max();
  for (auto& v : times) {
    if (v.size() < k) {
      continue;
    }
    std::nth_element(v.begin(), v.begin() + k, v.end());
    result = std::min(result, std::accumulate(v.begin(), v.begin() + k, 0u));
  }
  std::cout << result << '\n';
  return 0;
}