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 PYTHON3):

Solution:
from sys import stdin

inp = stdin.readline

n, k= [int(x) for x in inp().strip().split()]

a = [int(x) for x in inp().strip().split()]

a.sort()

d = {}

ans = -1
for i in a:
    j = 0
    while i != 0:
        if d.get(i,0) == 0:
            d[i] = [1, j]
        else:
            if d[i][0] < k:
                d[i][0] += 1
                d[i][1] += j
        if d[i][0] == k:
            if ans == -1:
                ans = d[i][1]
                break
            else:
                ans = min(ans, d[i][1])
                break
        i //= 2
        j += 1
print(ans)