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;
long long sol0(vector<long long> &a, int k) {
  int n = (int)a.size();
  long long res = 1LL << 55;
  long long psum = 0;
  for (int i = 0; i < n; i++) {
    if (i >= (k - 1)) {
      long long sum = a[i] * i - psum;
      long long v = sum - (i + 1 - k);
      res = min(res, v);
    }
    psum += a[i];
  }
  return res;
}
long long sol1(vector<long long> &a, int k) {
  int n = (int)a.size();
  long long res = 1LL << 55;
  long long psum = 0;
  long long ssum = 0;
  for (int i = 0; i < n; i++) ssum += a[i];
  for (int i = 0; i < n; i++) {
    ssum -= a[i];
    long long sum = (a[i] * i - psum) + (ssum - a[i] * (n - 1 - i));
    long long v = sum - (n - k);
    res = min(res, v);
    psum += a[i];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j < n && a[j] == a[i]) j++;
    int len = j - i;
    j--;
    if (len >= k) {
      cout << 0 << "\n";
      return 0;
    }
    i = j;
  }
  long long res = sol0(a, k);
  reverse(a.begin(), a.end());
  for (int i = 0; i < n; i++) a[i] *= -1;
  res = min(res, sol0(a, k));
  res = min(res, sol1(a, k));
  cout << res << "\n";
  return 0;
}