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 = 2e5 + 9, M = 1e5 + 9, OO = 0x3f3f3f3f, MOD = 1e9 + 7;
long long n, k;
long long arr[N];
map<long long, vector<long long> > canReach;
void process(long long f) {
  long long ret = 0;
  while (f) {
    long long accSum = 0;
    canReach[f].push_back(ret);
    f /= 2;
    ++ret;
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    process(arr[i]);
  }
  long long mn = OO;
  for (auto i : canReach) {
    sort(i.second.begin(), i.second.end());
    if (i.second.size() >= k) {
      long long tmp = 0;
      for (int j = 0; j < k; ++j) tmp += i.second[j];
      mn = min(mn, tmp);
    }
  }
  cout << mn;
}