When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?

You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.

A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Examples

Input

5
1 2 3 3 2


Output

4


Input

11
5 4 5 5 6 7 8 8 8 7 6


Output

5

Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
def main():
	n = int(input())
	L = [int(x) for x in input().split()]
	result = solver(L)
	print(result)

def solver(L):
	if len(L) <= 2:
		return len(L)
	else:
		lastSeen = dict()
		longest = 0
		lastSeen[L[0]] = 0
		lastSeen[L[1]] = 1
		for i in range(2, len(L)):
			x = L[i]
			lastSeen[x] = i
			plus1Index = -1
			minus1Index = -1
			plus2Index = -1
			minus2Index = -1
			if x + 1 in lastSeen:
				plus1Index = lastSeen[x + 1]
			if x - 1 in lastSeen:
				minus1Index = lastSeen[x - 1]
			if x + 2 in lastSeen:
				plus2Index = lastSeen[x + 2]
			if x - 2 in lastSeen:
				minus2Index = lastSeen[x - 2]
			min1Index = min(plus1Index, minus1Index)
			max2Index = max(plus2Index, minus2Index)
			if max2Index > min1Index:
				if i - max2Index > longest:
					longest = i - max2Index
			else:
				if i - min1Index > longest:
					longest = i - min1Index
			# biggestIndex = -1
			# if x - 1 in lastSeen and \
			# lastSeen[x - 1] > biggestIndex:
			# 	biggestIndex = lastSeen[x - 2]
			# if x + 1 in lastSeen and \
			# lastSeen[x + 1] > biggestIndex:
			# 	biggestIndex = lastSeen[x + 2]
			#if i - biggestIndex > longest:
			#	longest = i - biggestIndex
			if x - 3 in lastSeen:
				del lastSeen[x - 3]
			if x + 3 in lastSeen:
				del lastSeen[x + 3]	
		return longest

# def solver(L):
# 	longest = 0
# 	for i in range(len(L)):
# 		maximum = L[i]
# 		minimum = L[i]
# 		for j in range(i + 1, len(L)):
# 			if L[j] > maximum:
# 				maximum = L[j]
# 				if maximum - minimum > 1:
# 					if j - i > longest:
# 						longest = j - i
# 					break
# 			elif L[j] < minimum:
# 				minimum = L[j]
# 				if maximum - minimum > 1:
# 					if j - i > longest:
# 						longest = j - i
# 					break
# 		else:
# 			if len(L) - i > longest:
# 				longest = len(L) - i
# 	return longest

#L1 = [1, 2, 3, 3, 2]
#print(solver(L1))

#L2 = [5, 4, 5, 5, 6, 7, 8, 8, 8, 7, 6]
#print(solver(L2))
main()


