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 collections import defaultdict
from sys import stdin

input = stdin.readline

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

    dct = defaultdict(lambda: [])

    for x in arr:
        i = 0
        while x > 0:
            dct[x].append(i)
            x //= 2
            i += 1

    ndct = {}

    for key, v in dct.items():
        if len(v) >= k:
            v.sort()
            ndct[key] = sum(v[:k])

    print(min(v for _, v in ndct.items()))