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 long long pinf = 1e18 + 4;
const long long ninf = -pinf;
const long long N = 200005;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long A[n];
  long long S1[N];
  long long C[N];
  memset(C, 0, sizeof(C));
  memset(S1, 0, sizeof(S1));
  long long ans = INT_MAX;
  for (long long i = 0; i < n; i++) {
    cin >> A[i];
    C[A[i]]++;
    if (C[A[i]] == k) ans = 0;
  }
  for (long long j = 1; j < 18; j++) {
    for (long long i = 0; i < n; i++) {
      long long x = A[i] / 2;
      if (A[i] != x) {
        A[i] = x;
        S1[A[i]] += j;
        C[A[i]]++;
        if (C[A[i]] == k) ans = min(S1[A[i]], ans);
      }
    }
  }
  cout << endl;
  cout << ans << endl;
}