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;
long long n, m, mn = 999999999, x;
long long p[200009][20];
void solve() {
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    int z = 0;
    while (x > 0) {
      p[x][z]++;
      z++;
      x /= 2;
    }
  }
  for (long long i = 1; i < 200009; i++) {
    long long y = m, an = 0;
    for (long long j = 0; j < 20; j++) {
      if (p[i][j] < y)
        y -= p[i][j], an = an + j * p[i][j];
      else {
        an = an + j * y;
        y = 0;
        break;
      }
    }
    if (y == 0) mn = min(mn, an);
  }
  cout << mn;
}
int main() {
  long long t = 1;
  while (t--) solve();
}