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].
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def li(): return list(mi())

n=ii()
a=li()

minp=0;maxp=-1
mini,maxi=0,0
ans=0;curr=1

for i in range(1,n):
    if abs(a[i]-a[i-1])>1:
        minp=i
        maxp=-1
        ans=max(ans,curr)
        curr=1
        
    else:
        if maxp==-1:
            curr+=1
            
            if a[i]>=a[i-1]:
                maxp=i
                minp=i-1
                maxi=a[i]
                mini=a[i-1]
                
            else:
                maxp=i-1
                minp=i
                maxi=a[i-1]
                mini=a[i]
                
        else:
            if a[i]==maxi:
                maxp=i
                curr+=1
                
            elif a[i]==mini:
                minp=i
                curr+=1
                
            else:
                #tmaxi,tmini=maxi,mini
                if a[i]>maxi:
                    if maxi==mini:
                        curr+=1
                        maxi=a[i]
                        maxp=i
                        minp=i-1
                        
                    else:
                        ans=max(ans,curr)
                        curr=1+maxp-minp
                        maxp=i
                        minp=i-1
                        maxi=a[i];mini=a[i-1]
                    
                elif a[i]<mini:
                    if mini==maxi:
                        curr+=1
                        mini=a[i]
                        minp=i
                        maxp=i-1
                        
                    else:
                        ans=max(ans,curr)
                        curr=1+minp-maxp
                        maxp=i-1
                        minp=i
                        maxi=a[i-1];mini=a[i]

ans=max(ans,curr)                    
print(ans)