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() {
  int n, k;
  cin >> n >> k;
  vector<int64_t> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<int64_t> values;
  vector<int64_t> counts;
  int prev = 0;
  for (int i = 1; i < n; ++i) {
    if (a[i] != a[prev]) {
      values.push_back(a[prev]);
      counts.push_back(i - prev);
      prev = i;
    }
  }
  values.push_back(a[n - 1]);
  counts.push_back(n - prev);
  int m = values.size();
  vector<int64_t> prefCount(m, 0);
  vector<int64_t> prefSum(m, 0);
  for (int i = 1; i < m; ++i) {
    prefCount[i] = prefCount[i - 1] + counts[i - 1];
    prefSum[i] = prefSum[i - 1] + values[i - 1] * counts[i - 1];
  }
  vector<int64_t> sufCount(m, 0);
  vector<int64_t> sufSum(m, 0);
  for (int i = m - 2; i >= 0; --i) {
    sufCount[i] = sufCount[i + 1] + counts[i + 1];
    sufSum[i] = sufSum[i + 1] + values[i + 1] * counts[i + 1];
  }
  int64_t result = 1e18;
  for (int i = 0; i < m; ++i) {
    int64_t need = k - counts[i];
    if (need <= 0) {
      result = 0;
      continue;
    }
    {
      int64_t needl = min(need, prefCount[i]);
      int64_t costl =
          needl == 0 ? 0 : prefCount[i] * (values[i] - 1) - prefSum[i] + needl;
      int64_t needr = need - needl;
      int64_t costr =
          needr == 0 ? 0 : sufSum[i] - sufCount[i] * (values[i] + 1) + needr;
      result = min(result, costl + costr);
    }
    {
      int64_t needr = min(need, sufCount[i]);
      int64_t costr =
          needr == 0 ? 0 : sufSum[i] - sufCount[i] * (values[i] + 1) + needr;
      int64_t needl = need - needr;
      int64_t costl =
          needl == 0 ? 0 : prefCount[i] * (values[i] - 1) - prefSum[i] + needl;
      result = min(result, costl + costr);
    }
  }
  cout << result << endl;
  return 0;
}