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;
vector<long long> v[200001];
void solve() {
  long long x, y, t, ans = (int)1e9 + 7;
  cin >> x >> y;
  for (long long i = 0; i < x; ++i) {
    cin >> t;
    long long ct = 0;
    while (t) {
      v[t].push_back(ct);
      ct++;
      t /= 2;
    }
  }
  for (long long i = 0; i < 200001; ++i) {
    if (v[i].size() < y) continue;
    sort(v[i].begin(), v[i].end());
    long long sum = 0;
    for (long long k = 0; k < y; ++k) sum += v[i][k];
    ans = min(ans, sum);
  }
  cout << ans;
}
signed main() {
  long long t = 1;
  while (t--) solve();
  return 0;
}