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 readInts():
	return [int(x) for x in input().split()]


def minStepsForEqual(xs, k):
	poss = set()

	for x in xs:
		while x > 0:
			poss.add(x)
			x //= 2

	result = 10 ** 9

	for pos in poss:
		# print("For candidate = ", pos)
		cnt = []
		for x in xs:
			debug_x = x
			steps = 0
			while x > pos:
				x //= 2
				steps += 1
			if x == pos:
				# print("  ", debug_x, " needs ", steps, " steps")
				cnt.append(steps)
		if len(cnt) >= k:
			# print("  Count more than ", k, "!")
			cnt.sort()
			result = min(result, sum(cnt[:k]))
			# print("  Result = ", result)

	return result



if __name__ == "__main__":
	[N, k] = readInts()
	xs = readInts()
	result = minStepsForEqual(xs, k)
	print(str(result))