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:
n,k=[int(x) for x in input().split(' ')]
arr=sorted([int(x) for x in input().split(' ')])
ans=99999999999
cand=0
while cand <= arr[-1]:
	cnt=0
	cost=0
	for ind in range(n):
		op=0
		if(arr[ind]<cand):
			continue
		if(arr[ind]==cand):
			cnt+=1
			if(cnt==k):
				break
		else:
			temp=arr[ind]
			while(temp>cand):
				op+=1
				temp=temp//2
			if(temp==cand):
				cnt+=1
				cost+=op
				if(cnt==k):
					break
	# print(cand,cost,cnt)
	cand+=1
	if(cnt==k):
		ans=min(ans,cost)
print(ans)