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:
ll = lambda:list(map(int, input().split()))
testcases = 1
# [testcases] = ll()
def step(s,e):
	if s==e:
		return 0
	elif s<e:
		return -1
	count=0
	while(s>e):
		count+=1
		s = s//2
		if(s==e):
			return count
	return -1


for _ in range(testcases):
	[n,k] = ll()
	a =ll()
	a.sort()
	minres = 999999999
	t = a[-1]
	ix = -1
	for i in range(1,t+1):
		steps = []
		for j in a:
			x = step(j,i)
			if x!=-1:
				steps.append(x)
		if len(steps) < k:
			continue
		steps.sort()
		# print(steps,i)
		if minres > sum(steps[:k]):
			minres = sum(steps[:k])

			ix = i
	print(minres)