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].
import sys
def fastio():
    from io import StringIO
    from atexit import register
    global input
    sys.stdin = StringIO(sys.stdin.read())
    input = lambda : sys.stdin.readline().rstrip('\r\n')
    sys.stdout = StringIO()
    register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()

def debug(*var, sep = ' ', end = '\n'):
    print(*var, file=sys.stderr, end = end, sep = sep)

INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br

n, = I()
a = I()

ans = 0
j = 0
d = dd(int)
cur_min = cur_max = a[0]
for i in range(n):
    d[a[i]] += 1
    cur_min = min(cur_min, a[i])
    cur_max = max(cur_max, a[i])
    if cur_max - cur_min >= 2:
        if a[i] == cur_min:
            x = cur_max
        else:
            x = cur_min
        while d[x]:
            d[a[j]] -= 1
            j += 1
        d.pop(x)
    k = d.keys()
    cur_min = min(d.keys())
    cur_max = max(d.keys())
    ans = max(ans, i - j + 1)
print(ans)