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 debug import debug
from math import ceil, log2
import sys;input = sys.stdin.readline

inf = int(1e10)
n = int(input())
lis = list(map(int, input().split()))
# sparse table
k = int(log2(n))+2
tablemin = [[lis[i] if j == 0 else inf for j in range(k)] for i in range(n)]
tablemax = [[lis[i] if j == 0 else 0 for j in range(k)] for i in range(n)]
for j in range(1, k):
	for i in range(n-(1<<j)+1):
		tablemin[i][j] = min(tablemin[i][j-1], tablemin[i+(1<<(j-1))][j-1])
		tablemax[i][j] = max(tablemax[i][j-1], tablemax[i+(1<<(j-1))][j-1])
def qmax(a,b):
	s = log2(b-a+1)
	if ceil(s) == int(s): return tablemax[a][int(s)]
	s = int(s)
	return max(tablemax[a][s], tablemax[b-(1<<s)+1][s])
def qmin(a,b):
	s = log2(b-a+1)
	if ceil(s) == int(s): return tablemin[a][int(s)]
	s = int(s)
	return min(tablemin[a][s], tablemin[b-(1<<s)+1][s])

i, j = 0, 0
ans = 0
while j<n:
	ans = max(ans, j-i+1)
	if qmax(i, j+1) - qmin(i, j+1) < 2: j+=1
	else: i+=1
print(ans)

	



