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].
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=utf-8
#
# Copyright © 2015 vivek9ru <vivek9ru@Vivek9RU>
#
# Distributed under terms of the MIT license.

from math import acos,asin,atan,sqrt,pi,sin,cos,tan,ceil,floor
import re
from operator import itemgetter as ig

n=input()
lis=map(int,raw_input().split())
act=[lis[0]]
lact=[0]
ract=[0]
i=1
while i<n:
	while i<n and lis[i]==lis[i-1]:
		ract[-1]=i
		i+=1
	if i<n:
		act.append(lis[i])
		lact.append(i)
		ract.append(i)
		i+=1
l=len(act)
f=n
res=[0]*l
nex=[0]*l
res[l-1]=ract[l-1]-lact[l-1]+1
nex[l-1]=0
for i in range(l-2,-1,-1):
	nex[i]=act[i+1]-act[i]
	if i+2<l:
		a=act[i+1]-act[i]
		b=act[i+2]-act[i+1]
		if (a>0 and b>0) or (a<0 and b<0):
			res[i]=ract[i+1]-lact[i]+1
		elif (nex[i+2]==b):
			res[i]=ract[i+2]-lact[i]+1
		else:
			res[i]=res[i+2]+lact[i+2]-lact[i]
	else:
		res[i]=ract[i+1]-lact[i]+1
	#print i,res[i],ract[i],lact[i]
print max(res)
	
exit(0)
