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
n, m  = map(int, raw_input().split())

ar1 = map(int, raw_input().split())
mini = 100000000000000000000000
divi = {0: "k"}

for k in ar1:
	while k >= 1:
		divi[k] = "k"
		k /= 2
		divi[k] = "k"


for k in divi.keys():
	ar = []	
	for x in ar1:
		cont = 0
		fedeu = False
		while(x != k):
			if x < k:
				fedeu = True
				break
			
			x /= 2
			
			cont += 1
		
		if not fedeu:
			ar.append(cont)
	
	if len(ar) < m:
		continue
	ar.sort()
	suma = 0
	for i in xrange(m):
		suma += ar[i]
	mini = min(mini, suma)

print mini


	
