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:
n, k = [int(i) for i in input().split()]
lst = [int(i) for i in input().split()]


def hard(n, k, lst):
    a, result = [[] for i in range(1 + 2 * (10 ** 5))], 0
    for i in range(n):
        x, l = lst[i], 0
        while True:
            a[x].append(l)
            if x == 0:
                break
            x //= 2
            l += 1
    maximum = 1000000000
    for i in range(len(a)):
        a[i].sort()
        if len(a[i]) >= k:
            maximum = min(maximum, sum(a[i][:k]))
    return maximum


print(hard(n, k, lst))