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].
__author__ = 'cyborg'
from collections import defaultdict
def solve(a):
    minimum = maximum = a[0]
    cur_len = 1
    lar_len = 1
    stack  = defaultdict(int) # This will contain the latest continuous frequencies of maximum and minimum

    for i in xrange(1, len(a)):
        tod = []
        for s in stack:
            if s != maximum and s != minimum:
                tod.append(s)
        for s in tod:
            del stack[s]
        if cur_len > lar_len: lar_len = cur_len
        if a[i-1] == maximum:
            stack[maximum] += 1
        elif a[i-1] == minimum:
            stack[minimum] += 1
        if a[i-1] != maximum: stack[maximum] = 0
        if a[i-1] != minimum: stack[minimum] = 0

       # print stack
        if a[i] <= maximum and a[i] >= minimum:
            cur_len += 1
        elif a[i] == maximum+1 and minimum == maximum:
            maximum = a[i]
            cur_len += 1
        elif a[i] == minimum - 1 and minimum == maximum:
            minimum = a[i]
            cur_len += 1
        elif a[i] == maximum + 1 and minimum != maximum:
            maximum += 1
            if cur_len > lar_len: lar_len = cur_len
            minimum += 1
            cur_len = stack[minimum] + 1
        elif a[i] == minimum - 1 and minimum != maximum:
            minimum -= 1
            maximum -= 1
            if cur_len > lar_len : large_len = cur_len
            cur_len = stack[maximum] + 1
        #print a[i],'cur_len',cur_len,'maximum',maximum,'minimum',minimum,'stack', stack
    return max(cur_len, lar_len)
n=input()
print solve(map(int, raw_input().split()))
