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 = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
d = dict()
c = 0
f = 0
a = [0 for i in range(200001)]
min1 = -1
while(l[-1] != 0):
	for i in range(f,len(l)):
		if l[i] in d:
			d[l[i]] += 1
		else:
			d[l[i]] = 1
		a[l[i]] += c
		if(d[l[i]] >= k):
			#print(min1,l[i],d[l[i]],a[l[i]])
			if(min1 == -1):
				min1 = a[l[i]]
			else:
				min1 = min(min1,a[l[i]])
		if(l[i] == 0):
			f = i+1
		l[i] = l[i]//2
	#print(l)
	c += 1
print(min1)