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 = int(raw_input())
a = map(int, raw_input().split())

log2 = [0] * (n + 1)
for i in xrange(2, n + 1):
    log2[i] = log2[i >> 1] + 1
mins = [a[::]]
maxes = [a[::]]
levels = log2[n]
length = 1
for level in xrange(1, levels + 1):
    mins.append(mins[-1][::])
    maxes.append(maxes[-1][::])
    i = 0
    while i + length < n:
        mins[level][i] = min(mins[level][i], mins[level][i + length])
        maxes[level][i] = max(maxes[level][i], maxes[level][i + length])
        i += 1
    length <<= 1

def rangeMin(left, right):
    length = right - left + 1
    level = log2[length]
    return min(mins[level][left], mins[level][right - (1 << level) + 1])

def rangeMax(left, right):
    length = right - left + 1
    level = log2[length]
    return max(maxes[level][left], maxes[level][right - (1 << level) + 1])

ret = 0
left = right = 0
while right < n:
    while right < n and rangeMax(left, right) - rangeMin(left, right) <= 1:
        right += 1
    ret = max(ret, right - left)
    while right < n and left <= right and rangeMax(left, right) - rangeMin(left, right) > 1:
        left += 1
print ret
