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:
def get(a):
	return max([a.count(i) for i in set(a)])


n, k = map(int, input().split())
a = sorted(list(map(int, input().split())))
b = a.copy()

can = set()
can.add(0)
for i in range(n):
	while b[i] != 0:
		can.add(b[i])
		b[i] //= 2

can = list(can)
ans = []
for i in range(len(can)):
	b = a.copy()
	local_ans = 0
	count = 0
	for j in range(n):
		now_count = 0
		while b[j] > can[i]:
			b[j] //= 2
			now_count += 1
		if b[j] == can[i]:
			count += now_count
			local_ans += 1
		if local_ans >= k:
			break
	if local_ans >= k:
		ans.append(count)

# print(ans)
print(min(ans))