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 sys import stdin,stdout
from math import log2
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
def RMinQ(low,high):
    l=(high-low)+1
    k=int(log2(l))
    return min(a[sp[low][k]],a[sp[low+l-(1<<k)][k]])
def RMaxQ(low,high):
    l=(high-low)+1
    k=int(log2(l))
    return max(a[sp1[low][k]],a[sp1[low+l-(1<<k)][k]])
for _ in range(1):#nmbr()):
    n=nmbr()
    a=lst()
    col=int(log2(n))+1
    sp=[[0 for _ in range(col)]for _ in range(n)]
    sp1=[[0 for _ in range(col)]for _ in range(n)]
    for i in range(n):
        sp[i][0]=sp1[i][0]=i
    j=1
    while (1<<j)<=n:
        i=0
        while (i+(1<<j)-1)<n:
            if a[sp[i][j-1]]<a[sp[i+(1<<(j-1))][j-1]]:
                sp[i][j]=sp[i][j-1]
            else:sp[i][j]=sp[i+(1<<(j-1))][j-1]
            i+=1
        j+=1
    j=1
    while (1<<j)<=n:
        i=0
        while (i+(1<<j)-1)<n:
            if a[sp1[i][j-1]]>a[sp1[i+(1<<(j-1))][j-1]]:
                sp1[i][j]=sp1[i][j-1]
            else:sp1[i][j]=sp1[i+(1<<(j-1))][j-1]
            i+=1
        j+=1
    ans=0
    for i in range(n):
        l=i;r=n-1
        while l<=r:
            mid=(l+r)>>1
            if abs(RMaxQ(i,mid)-RMinQ(i,mid))<=1:l=mid+1
            else:r=mid-1
        ans=max(ans,r-i+1)
    print(ans)