【CodeForces 602B】G

Description

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, ..., a_n. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |a_i + 1 - a_i| ≤ 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 a_i 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, ..., a_n (1 ≤ a_i ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Sample Input

Input

5
1 2 3 3 2

Output

4

Input

11
5 4 5 5 6 7 8 8 8 7 6

Output

5

Hint

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].

读入数据时如果当前的和之前的相同则记录，处理时维护两个数字，题意的一个序列里最多两个不同数字，如果不符合就跳出，符合则判断下一个（j=b[j])

#include<stdio.h>
#include<algorithm>
using namespace std;

long long n,ans,len,cont,num1,num2=100005,a[100005],b[100005];
int main()
{
    scanf("%lld",&n);
    int i,j;
    for(i=1; i<=n; i++)
    {
        scanf("%lld",&a[i]);
        if(a[i]==a[i-1])
        {
            if(cont==0) //cont=0前一个不重复
                cont=b[i-1];
            b[i]=cont;
        }
        else
            b[i]=i-1;
        cont=0;
    }
    for(i=n; i>0; i--)
    {
        num1=a[i];
        num2=100005;
        len=1;
        j=i-1;
        while(j>0)
        {
            if(num1==a[j]||num2==a[j])
            {
                len+=j-b[j];
                j=b[j];
                continue;
            }
            else if(num2==100005)
            {
                num2=a[j];
                len+=j-b[j];
                j=b[j];
            }
            else break;
        }
        ans=max(ans,len);
        if(i<ans)break;
    }
    printf("%lld
",ans);
    return 0;
}