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;
vector<long long> dp[200007];
map<long long, long long> mp;
vector<long long> un;
void make_di(long long n, long long c) {
  dp[n].push_back(c);
  if (mp[n] == 0) un.push_back(n);
  mp[n] = 1;
  if (n == 0) return;
  make_di(n / 2, c + 1);
}
int main() {
  long long i, j, a, n, k, ans = 1e9, sum;
  vector<long long> v;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
    make_di(a, 0);
  }
  for (i = 0; i < 200007; i++) {
    if (dp[i].size() >= k) sort(dp[i].begin(), dp[i].end());
  }
  for (i = 0; i < un.size(); i++) {
    a = un[i];
    if (dp[a].size() < k) continue;
    sum = 0;
    for (j = 0; j < k; j++) {
      sum += dp[a][j];
    }
    ans = min(sum, ans);
  }
  cout << ans << "\n";
  return 0;
}