B. Approximating a Constant Range

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

Copy

5
1 2 3 3 2

output

Copy

4

input

Copy

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

output

Copy

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].

题意:

给一列数，求最小值最大值的差值不超过1的最大区间长度。

思路：解法很多rmq，线段树，都可以我写的是单调队列写的

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
#define LL long long

using namespace std;
const int maxn=1e5+100;
int a[maxn];

int main()
{
    deque<int>Q1;
    deque<int>Q2;
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    int pre=1;
    int ans=-1;
    for(int i=1;i<=n;i++)
    {
        while(!Q1.empty()&&Q1.back()<a[i])
            Q1.pop_back();
        while(!Q2.empty()&&Q2.back()>a[i])
        {
            Q2.pop_back();
        }
        Q1.push_back(a[i]);
        Q2.push_back(a[i]);
        while(Q1.front()-Q2.front()>1)
        {
            if(a[pre]==Q1.front())
                Q1.pop_front();
            if(a[pre]==Q2.front())
                Q2.pop_front();
            pre++;
        }
        ans=max(ans,i-pre+1);
    }
    printf("%d\n",ans);
    return 0;
}