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].
#####################
#1: Array, 
#R -> ++Count[]
#L -> --Count[]
# line1 = input().split(" ")
# arr = list(map(int,input().split()))
# n = int(line1[0])
# k = int(line1[1])
# counts = {}
# l = 0
# r = 0
# count = 0
# for i in range(len(arr)):
#     if arr[i] in counts:
#         counts[arr[i]] += 1
#     else:
#         count += 1
#         counts[arr[i]] = 1
#         if count >= k:
#             r = i
#             break
# for i in range(len(arr)):
#     l = i
#     counts[arr[i]] -= 1
#     if counts[arr[i]] == 0:
#         l = i
#         break
# if count < k:
#     print("-1 -1")
# else:
#     print(l + 1, r + 1)

#2: George and Round
# n, m = map(int, input().split())
# arrA = list(map(int, input().split()))
# arrB = list(map(int, input().split()))
# k = n
# pA = 0
# pB = 0
# while pB < m:
#     if arrB[pB] >= arrA[pA]:
#         k -= 1
#         pA += 1
#         if pA >= n:
#             break
#     pB += 1
# print(k)

#3
# n, t = map(int, input().split())
# a = list(map(int, input().split()))
# l = 0
# r = 0
# maxx = 0
# summ = 0
# while r <= n:
#     if r == n:
#         maxx = max(maxx, (r - 1) - l + 1)
#         break
#     if summ + a[r] > t:
#         maxx = max(maxx, (r - 1) - l + 1)
#         summ -= a[l]
#         l += 1
#     else:
#         summ += a[r]
#         r += 1
# print(maxx)

#6: 
#each segment must have 2 distinct number
#val 1, val 2 without map or dictionary
n = int(input())
a = list(map(int, input().split()))
fre = [0] * (10 ** 5 + 5)
diff = 0
j = 0
longest_range = 0
for i in range(n):
    if fre[a[i]] == 0:
        diff += 1
    fre[a[i]] += 1
 
    while j < n and diff > 2:
        if fre[a[j]] == 1:
            diff -= 1
        fre[a[j]] -= 1
        j += 1
     
    longest_range = max(longest_range, i - j + 1)
 
print(longest_range)

#7: https://codeforces.com/contest/892/problem/B
# n = int(input())
# a = list(map(int, input()))


#8: https://codeforces.com/problemset/problem/6/C
# n = int(input())
# a = list(map(int, input().split()))
# l = 0
# r = n - 1
# alice = 0
# bob = 0
# tA = 0
# tB = 0
# #<=0: alice turn, >0 bob turn (<= 0 cuz bob is a boy :v)
# while l <= r:
#     if tA - tB <= 0:
#         alice += 1
#         tA += a[l]
#         l += 1
#     else:
#         bob += 1
#         tB += a[r]
#         r -= 1
# print(alice, bob)