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].
#import random
n = int(raw_input())
a = map(int, raw_input().split())
#r = [random.randint(-1, 1) for i in xrange(n)]
#a = [sum(r[:i]) for i in xrange(n)]
st = 0
c = 1
best = 1
mn = a[0]
mx = a[0]
lmn = 0
lmx = 0
#print a
while c < n:
	if a[c] - mn > 1 or mx - a[c] > 1:
		best = max(best, c-st)
		if a[c] > mx:
			st = lmn + 1
			lmn = c-1
			lmx = c
			mn = a[c]-1
			mx = a[c]
		else:
			st = lmx+1
			lmn = c
			lmx = c-1
			mn = a[c]
			mx = a[c]+1
	if a[c] <= mn:
		mn = a[c]
		lmn = c
	if a[c] >= mx:
		mx = a[c]
		lmx = c
	c += 1
best = max(best, n-st)
print best

#bt = 0
#bi, bj = 0, 0
#for i in xrange(0, n-1):
#	for j in xrange(i+1, n):
#		if max(a[i:j]) - min(a[i:j]) < 2:
#			if j-i > bt:
#				bt = j-i
#				bi = i
#				bj = j
#print bt, bi, bj