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
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  unordered_map<int, map<int, int> > m1;
  unordered_map<int, int> m2;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int numSteps = 0;
    while (a[i] > 0) {
      m2[a[i]]++;
      m1[a[i]][numSteps]++;
      a[i] /= 2;
      ++numSteps;
    }
  }
  long long target = 0;
  long long steps = INT_MAX;
  for (auto it : m2) {
    if (it.second >= k) {
      long long numSteps = 0;
      int nums = 0;
      auto it1 = m1[it.first].begin();
      while (nums < k) {
        long long tmp = min(k - nums, it1->second);
        numSteps += (tmp * it1->first);
        nums += tmp;
        it1++;
      }
      if (numSteps < steps) {
        steps = numSteps;
        target = it.first;
      }
    }
  }
  cout << steps << endl;
  return 0;
}
