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].
from pprint import pprint as pp
def GI(): return int(raw_input())
def GIS(): return map(int, raw_input().split())

def main():
    n = GI()
    xs = GIS()
    
    ml = 0
    minv, maxv = xs[0:1] * 2
    minp, maxp = 0, 0

    for p, x in enumerate(xs[1:], 1):
        c = 1 if p > 1 else 2
        if x > maxv:
            c -= 1
            minv, minp = maxv, maxp
            maxv, maxp = x, p
        elif x < minv:
            maxv, maxp = minv, minp
            minv, minp = x, p
            c -= 1
        ml = max(ml, (p+c - minp), (p+c - maxp))

    return ml

def main():
    n = GI()
    xs = GIS()
    d = {}
    ml = 0
    get = lambda a: d.get(a, -1)

    for p, x in enumerate(xs):
        d[x] = p

        mo = get(x-1)
        po = get(x+1)
        mt = get(x-2)
        pt = get(x+2)
        l = [mo, po, mt, pt]
        l.sort()
        ml = max(ml, p-l[-2])

    return ml

print main()
