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].

datain = True

if datain:
    n = int(input())
    a = list(map(int, input().split()))
else:
    n = 5    
    a = [1, 2, 3, 3, 2]

sto = {}
le  = 0
ri  = 0
sa  = len(a)
cnt = 0

# Number of different elements does not exceed 2 :)
# O(2*n)
for ri in range(sa):
	if not (a[ri] in sto):
		sto[a[ri]] = 1
	else:
		sto[a[ri]] += 1
		
	while (le < sa) and (len(sto) > 2):
		if (sto[a[le]] == 1):
			sto.pop(a[le])
		else:
			sto[a[le]] -= 1
			
		le += 1
	
	cnt = max(cnt, ri - le + 1)
	
print(cnt)





"""
# O(n**2)
sa   = len(a)
minv = a[0]
maxv = a[0]
le   = 0
ri   = -1
numb = 0
fina = 0

for le in range(sa):
    ri   = le - 1
    numb = 0
    minv = a[le]
    maxv = a[le]
    
    while (ri+1 < sa) and (abs(a[ri+1]-minv) <= 1) and (abs(a[ri+1]-maxv) <= 1):
        numb += 1
        ri   += 1
        
        minv  = min(minv, a[ri])
        maxv  = max(maxv, a[ri])
            
    fina = max(fina, numb)
    
print(fina)
"""