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 long long INF = 1e16;
const double EPS = 1e-9;
const long long MOD = (long long)(1e9 + 7);
const long long MAXV = (long long)(2e5 + 10);
const long long MAXE = (long long)(1e6 + 10);
int n, k;
int A[MAXV];
vector<int> ans[MAXV];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    int curr = 0;
    int aux = A[i];
    while (1) {
      ans[aux].push_back(curr);
      if (aux == 0) break;
      aux /= 2;
      curr++;
    }
  }
  long long sol = INF;
  for (int i = 0; i < MAXV; i++) {
    if (ans[i].size() < k) continue;
    sort(ans[i].begin(), ans[i].end());
    long long h = 0;
    for (int j = 0; j < k; j++) h += ans[i][j];
    sol = min(sol, h);
  }
  cout << sol << "\n";
  return 0;
}
