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 = 20000005;
int n, k, x, t, a, b, m;
string s;
vector<int> v;
void solve() {
  cin >> n >> k;
  map<int, pair<int, int>> cnt;
  for (int i = 0; i < n; i++) {
    cin >> x;
    v.push_back(x);
    cnt[x] = {cnt[x].first + 1, 0};
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    int num = v[i];
    int cur = 0;
    while (num > 0) {
      cur++;
      num /= 2;
      int count = cnt[num].first;
      int moves = cnt[num].second;
      if (count < k) cnt[num] = {count + 1, moves + cur};
    }
  }
  int mn = 1e9;
  for (auto m : cnt) {
    if (m.second.first >= k) mn = min(m.second.second, mn);
  }
  cout << mn;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}