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;
struct Edge {
  int from, to, dist;
  Edge(int u, int v, int d) : from(u), to(v), dist(d) {}
};
int n, k;
int a[100007 * 2];
set<int> all;
map<int, long long> mp, ans;
long long anss = 1000000007;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    int tmp = a[i], cnt = 0;
    while (tmp != -1) {
      mp[tmp]++;
      ans[tmp] += cnt;
      if (mp[tmp] == k) {
        anss = min(anss, ans[tmp]);
      }
      if (tmp != 0)
        tmp >>= 1;
      else
        tmp = -1;
      cnt++;
    }
  }
  cout << anss << endl;
  return 0;
}