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;
const int N = 1e6, INF = 1e9 + 7;
int n, k, a[N];
vector<pair<int, int> > g;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    g.push_back(make_pair(a[i], 0));
    int c = 1;
    while (a[i] > 0) {
      g.push_back(make_pair(a[i] / 2, c));
      c++, a[i] /= 2;
    }
  }
  int ans = INF;
  sort(g.begin(), g.end());
  for (int i = 0, sum = 0, j = 0; i < (int)g.size(); i++) {
    if (i == 0 || g[i].first != g[i - 1].first) {
      sum = g[i].second, j = 1;
    } else {
      j += 1, sum += g[i].second;
    }
    if (j == k) {
      ans = min(ans, sum);
    }
  }
  cout << ans;
}