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
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
const long double PI = (long double)(3.1415926535897932384626433832795);
void solve() {
  map<int, vector<int> > mp;
  int n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; ++i) {
    int tmp;
    cin >> tmp;
    int cnt = 0;
    mp[tmp].push_back(cnt);
    while (tmp) {
      tmp /= 2;
      cnt++;
      mp[tmp].push_back(cnt);
    }
  }
  long long mn = (long long)1e18;
  for (auto x : mp) {
    if (x.second.size() >= k) {
      long long tmp = 0;
      sort((x.second).begin(), (x.second).end());
      for (long long i = 0; i < k; ++i) {
        tmp += x.second[i];
      }
      mn = min(mn, tmp);
    }
  }
  cout << mn << endl;
}
int main() {
  long long t = 1;
  for (long long i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}
