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 int binaryExponentiation(long long int x, long long int n) {
  if (x == 0) return 0;
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  if (x == 0) return 0;
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  else
    return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
bool prime_check(long long x) {
  bool prime = (x >= 2);
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      prime = false;
      break;
    }
  }
  return prime;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long arr[n];
  long long val[n];
  for (long long int i = 1; i < n + 1; i++) {
    cin >> arr[i - 1];
  }
  sort(arr, arr + n);
  long long mini = 1000000000000000000;
  for (long long int i = 0; i < arr[n - 1] + 1; i++) {
    long long s = 0;
    long long c = 0;
    for (long long int j = 0; j < n; j++) {
      c = 0;
      long long x = arr[j];
      if (x < i)
        val[j] = INT_MAX;
      else {
        while (x >= i) {
          if (x == i) {
            val[j] = c;
            break;
          }
          c++;
          x /= 2;
          if (x < i) val[j] = INT_MAX;
        }
      }
    }
    sort(val, val + n);
    for (long long int j = 0; j < k; j++) {
      s += val[j];
    }
    mini = min(mini, s);
  }
  cout << mini << "\n";
  return 0;
}