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 longest(m):
    if len(m) <= 1:
        return len(m)
    min = m[0]
    max = m[0]
    count = 1
    mm = [1 for i in m]
    result = 1
    for idx, _ in enumerate(m):
        if idx == 0:
            continue
        if m[idx] == m[idx - 1]:
            mm[idx] = mm[idx-1] + 1
            mm[idx - 1] = 0
    for idx, num in enumerate(m[1:]):
        # print count, result, idx + 1, max, min
        if abs(min - num) == 2 or abs(max - num) == 2:
            count = 1 + mm[idx]
            if count > result:
                result = count
            if num > m[idx]:
                max = num
                min = m[idx]
            else:
                max = m[idx]
                min = num
        else:
            if min > num:
                min = num
            if max < num:
                max = num
            count += 1
            if count > result:
                result = count
    return result    

from sys import stdin
#inp = open("in", "r")
inp = stdin
n = int(inp.readline().strip())
if n > 1:
    m = map(int, inp.readline().strip().split())
    print longest(m[:n])
else:
    print n 
