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;
const int mx = 1e5 + 100;
const int md = 1000000007;
priority_queue<int, vector<int>, greater<int> > pq;
bool compare(int &a, int &b) { return a < b; }
vector<priority_queue<int, vector<int> > > arr(mx * 2);
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i += 1) {
    int x, z = 0;
    scanf("%d", &x);
    while (true) {
      arr[x].push(z);
      if (arr[x].size() > m) arr[x].pop();
      if (x == 0) break;
      z++;
      x /= 2;
    }
  }
  long long ans = -1, temp = 0;
  for (int i = 0; i < mx * 2; i += 1) {
    temp = 0;
    if (arr[i].size() >= m) {
      for (int j = 0; j < m; j += 1) {
        temp += arr[i].top();
        arr[i].pop();
      }
      if (ans == -1) ans = temp;
      ans = min(ans, temp);
    }
  }
  cout << ans << endl;
}
