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:
import math


def cnt(n, x):
    res = 0
    while n > x:
        n //= 2
        res += 1
    if n == x:
        return res
    return -1


def main():
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = set()
    b.add(0)
    for i in a:
        while i > 0:
            b.add(i)
            i //= 2
    ans = int(10e9)
    for x in b:
        s = []
        for i in a:
            c = cnt(i, x)
            if c >= 0:
                s.append(c)
        s.sort()
        if len(s) >= k:
            l = sum(s[:k])
            ans = min(ans, l)
    print(ans)

main()