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=input()
lis=map(int,raw_input().split())

currentactual=[lis[0]]
previousread=[0]
finalread=[0]
finalread2=[0]
previousread2=[0]
i=1

while i<n:
        #print 1
	while i<n and lis[i]==lis[i-1]:
		finalread[-1]=i
                finalread2[-1]=i
                #print i
		i+=1
	if i<n:
		currentactual.append(lis[i])
		previousread.append(i)
		finalread.append(i)
		i+=1

l=len(currentactual)
l2=len(currentactual)

f=n
f2=n

res=[0]*(l+1)
nextar=[0]*(l+1)
res[l-1]=finalread[l-1]-previousread[l-1]+1
nextar[l-1]=0

for i in range(l-2,-1,-1):
    nextar[i]=currentactual[i+1]-currentactual[i]

for i in range(l-2,-1,-1):
	if i+2<l:
		k1=currentactual[i+1]-currentactual[i]
		k2=currentactual[i+2]-currentactual[i+1]
		if (k1>0 and k2>0):
			res[i]=finalread[i+1]-previousread[i]+1
		elif (k1<0 and k2<0):
			res[i]=finalread[i+1]-previousread[i]+1
		elif (nextar[i+2]==k2):
			res[i]=finalread[i+2]-previousread[i]+1
		else:
			res[i]=res[i+2]+previousread[i+2]-previousread[i]
	else:
		res[i]=finalread[i+1]-previousread[i]+1
ans=0
for i in res:
    ans=max(ans,i)

print ans
