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 i1, i2, i3, i4, t1, t2, t3, l1, ans;
long long n, k, ac[200050][25], s[200050], c[200050], total[200050];
vector<long long> v;
int main() {
  cin >> n >> k;
  for (i1 = 0; i1 < n; i1++) {
    for (i2 = 0; i2 < 25; i2++) ac[i1][i2] = -1;
  }
  for (i1 = 1; i1 <= n; i1++) {
    cin >> t1;
    v.push_back(t1);
  }
  sort(v.begin(), v.end());
  for (i1 = 0; i1 < n; i1++) {
    t1 = v[i1];
    s[i1] = round(log(t1) / log(2) + 0.5);
    for (i2 = s[i1]; i2 >= 0; i2--) {
      ac[i1][i2] = t1;
      t1 /= 2;
    }
  }
  for (i1 = 0; i1 < n; i1++) {
    for (i2 = 0; i2 < 25; i2++) {
      if (ac[i1][i2] == -1) continue;
      c[ac[i1][i2]]++;
      if (c[ac[i1][i2]] <= k) total[ac[i1][i2]] += s[i1] - i2;
    }
  }
  ans = 100000000;
  for (i1 = 1; i1 < 200050; i1++) {
    if (c[i1] >= k) ans = min(ans, total[i1]);
  }
  cout << ans;
  return 0;
}