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;
int n, k, t1, t2, z, mn = 0x3f3f3f3f;
vector<int> v;
map<int, int> cnt;
map<int, vector<int> > cost;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> t1;
    t2 = 0;
    while (t1) {
      cnt[t1]++;
      if (cnt[t1] == 1) {
        v.push_back(t1);
      }
      cost[t1].push_back(t2);
      t1 >>= 1, t2++;
    }
    cost[0].push_back(t2);
  }
  v.push_back(0);
  for (int i = 0; i < v.size(); i++) {
    if (cnt[v[i]] >= k) {
      t1 = 0;
      sort(cost[v[i]].begin(), cost[v[i]].end());
      for (int j = 0; j < k; j++) {
        t1 += cost[v[i]][j];
      }
      mn = min(mn, t1);
    }
  }
  cout << mn << endl;
}