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__ = 'trunghieu11'

def calc(n, A, val):
    answer = 0
    counter = 0
    sameCounter = 0
    cur = A[0]
    same = A[0]
    i = 0
    while i < len(A):
        if A[i] == cur or A[i] + val == cur:
            counter += 1
        else:
            answer = max(answer, counter)
            if i - 1 >= 0 and A[i - 1] + val == A[i]:
                counter = sameCounter + 1
                cur = A[i]
            else:
                counter = 1
                cur = A[i]

        if same == A[i]:
            sameCounter += 1
        else:
            same = A[i]
            sameCounter = 1
        i += 1

    return max(answer, counter)

def solve(n, A):
    return max(calc(n, A, 1), calc(n, A, -1))

if __name__ == '__main__':
    n = int(raw_input())
    A = map(int, raw_input().split(" "))
    print solve(n, A)