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 math
n,m=map(int,input().split())
l=list(map(int,input().split()))
a=sorted(l)
mans=10000000
for k in range(0,n-m+1):
	x=int(math.log(a[k],2))
	for i in range(x+2):
		ans=i
		equal=(a[k]//(2**i))
		count=1
		flag=0
		#print (ans,equal)
		for j in range(k+1,n):
			temp=a[j]
			op=0
			#print (k,count,temp,equal)
			if count==m:
				break
			else:
				while temp>equal:
					temp=temp//2
					op+=1
				if temp==equal:
					count+=1
					ans+=op
			#print (count,equal,ans)
		if count!=m:
			flag=1
		if flag==0:
			if ans<mans:
				mans=ans
	#print (k,ans,flag)
print (mans)