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 = int(input())
a = list(map(int,input().split()))
k,l = a[0],a[1]
if k==l:
    base = k
    case = 0
else:
    case =1
    base = max(k,l)
    mn = min(k,l)
    lmn = 0 if k==mn else 1
    lmx = 0 if lmn else 1
l = 2
best =2
for i in range(2,n):
    best = max(best,l)
    if case == 1:
        if a[i]>base:
            l = i-lmn
            mn = base
            base = a[i]
            lmn = i-1
            lmx = i
        elif a[i]==base:
            l+=1
            lmx = i
        elif a[i]==mn:
            l+=1
            lmn = i
        else:
            base = mn
            mn = a[i]
            l = i-lmx
            lmx = i-1
            lmn = i
    else:
        l+=1
        if a[i]!=base:
            case =1
            mn = min(a[i],base)
            base = max(a[i], base)
            lmn = i if a[i]==mn else i-1
            lmx = i if a[i]==base else i-1

print(max(best,l))