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].
N = int(raw_input())
a = map(int, raw_input().split())

L = 0
R = 1
_max = max(a[0], a[1])
_min = min(a[0], a[1])
# minc and maxc here is WRONG, update it latter when we found different (case 1 1 1 2 3)
maxc = 1
minc = 1

res = 2
while R < N - 1:
	R += 1
	if a[R] > _max:
		if _max - _min == 1:
			# run if we has different min and max
			while minc > 0 and L < N:
				# trim left side until there is no min element left
				if a[L] == _min:
					minc -= 1
				if a[L] == _max:
					maxc -= 1
				L += 1
			# switch min and max -> so switch minc and maxc also
			minc = maxc
			# new max element has count 1
			maxc = 1
			_min += 1
		else:
			# update case (1 1 1 2 3) here
			minc = minc + maxc
			maxc = 1
		_max = a[R]
	elif a[R] == _max:
		maxc += 1
	elif a[R] < _min:
		if _max - _min == 1:
			# run if we has different min and max
			while maxc > 0 and L < N:
				# trim left side until there is no max element left
				if a[L] == _max:
					maxc -= 1
				if a[L] == _min:
					minc -= 1
				L += 1
			# switch max and min -> so switch maxc and minc also
			maxc = minc
			# new min element has count 1
			minc = 1
			_max -= 1
		else:
			# update case (1 1 1 2 3) here
			maxc = maxc + minc
			minc = 1		
		_min = a[R]
	elif a[R] == _min:
		minc += 1
	
	if R - L + 1 > res:
		res = R - L + 1

print(res)
