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>
constexpr long long MAX_RES = 1LL << 62;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int n, k;
  std::cin >> n >> k;
  std::vector<long long> a(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
  }
  std::sort(a.begin(), a.end());
  std::vector<long long> b(n);
  std::partial_sum(a.begin(), a.end(), b.begin());
  long long res = MAX_RES;
  for (int l = 0, r = 0; l < n;) {
    while (r < n && a[r] == a[l]) {
      ++r;
    }
    int numCnt = r - l;
    if (numCnt >= k) {
      res = 0;
    } else {
      int numLeft = l;
      long long costLeft = 0;
      if (numLeft) {
        costLeft = (1LL * numLeft * (a[l] - 1)) - b[l - 1];
      }
      int numRight = n - r;
      long long costRight = 0;
      if (numRight) {
        costRight = (b[n - 1] - b[r - 1] - 1LL * numRight * (a[l] + 1));
      }
      if (numLeft + numCnt >= k || numRight + numCnt >= k) {
        if (numLeft + numCnt >= k) {
          res = std::min(res, costLeft + k - numCnt);
        }
        if (numRight + numCnt >= k) {
          res = std::min(res, costRight + k - numCnt);
        }
      } else {
        res = std::min(res, costLeft + costRight + k - numCnt);
      }
    }
    l = r;
  }
  std::cout << res << "\n";
  return 0;
}