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 = input()
arr = map(int,raw_input().split())
minm = [min(arr[0],arr[1]),arr.index(min(arr[0],arr[1]))]
maxm = [max(arr[0],arr[1]),arr.index(max(arr[0],arr[1]))]
maxsize = 0
i,j = 0,1
while i < n and j < n:
    while maxm[0] - minm[0] > 1:
        if maxm[1] < minm[1]:
            i= maxm[1] + 1
            maxm[1] = i
            maxm[0] = arr[i]
        else:
            i = minm[1] + 1
            minm[1] = i
            minm[0] = arr[i]
        if i == j:
            j = i + 1
    if arr[j] >= maxm[0]:
        maxm[0] = arr[j]
        maxm[1] = j
    if arr[j] <= minm[0]:
        minm[0] = arr[j]
        minm[1] = j
    if maxm[0] - minm[0] <= 1:
        maxsize = max(abs(i-j),maxsize)
        j += 1
print maxsize+1