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].
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
z,zz=input,lambda:list(map(int,z().split()))
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from re import *
from sys import *
from math import *
from heapq import *
from queue import *
from bisect import *
from string import *
from itertools import *
from collections import *
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from collections import Counter as cc
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def output(answer):stdout.write(str(answer))
###########################---Test-Case---#################################
"""

  If you Know me , Then you probably don't know me !


"""
###########################---START-CODING---##############################


n=int(z())
arr=zzz()

dp=[0]*100010
ans=0

for num in arr:
    dp[num] +=1
    dp[num-1] +=1
    dp[num+1] =0
    if num!=1:
        dp[num-2]=0
    ans=max(ans,dp[num],dp[num-1])
print(ans)













