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
from itertools import islice

if __name__ == '__main__':
    _, k = tuple(map(int, input().split()))
    arr = list(map(int, input().split()))

    counts = [None] * (2 * (10 ** 5) + 1)

    def append_counts(xx, cnt):
        if counts[xx] is None:
            counts[xx] = []
        counts[xx].append(cnt)

    for a in arr:
        cur = 0
        x = a

        while x > 0:
            append_counts(x, cur)
            x //= 2
            cur += 1

    result = 2 * (10 ** 7)
    for i in counts:
        if i and len(i) >= k:
            i.sort()
            result = min(result, sum(islice(i, 0, k)))

    print(result)




