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 sys
from collections import defaultdict

def main():
    n, k = [int(x) for x in sys.stdin.readline().split(" ")]
    arr = [int(x) for x in sys.stdin.readline().split(" ")]
    # vals[x][a] holds number of iterations to get from x to a
    arr.sort()
    vals = defaultdict(list)
    for i, a in enumerate(arr):
        count = 0
        while (a > 0):
            vals[a].append(count)
            a = a // 2
            count += 1
    minCount = 99999999999999
    for x,v in vals.items():
        if len(v) < k:
            continue
        else:
            a = sorted(v)
            minCount = min(minCount, sum(a[:k]))
    return minCount

print(main())