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 *
n,k = [int(i) for i in input().split()]
A = [int(i) for i in input().split()]
A.sort()
d = defaultdict(lambda: [0,0])
for i in A:
	num = i
	stp = 0
	while num>0:
		if d[num][0]<k:
			d[num][0]+=1
			d[num][1]+=stp
		num = num//2
		stp +=1
	if d[num][0]<k:
			d[num][0]+=1
			d[num][1]+=stp

mn = 10**20
for i in d:
	if(d[i][0]>=k):
		mn = min(d[i][1],mn)

print(mn)