CF-Approximating a Constant Range

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].

---

思路：

一开始各种想LIS，但是CF再次证明了前两道题目全是想法题（虽然想法并不比算法简单= =~）

首先是要在一整个数段中找到其中的一个子数段，在它的M-m<=1的情况下使其len最大

那么这里就涉及到两个思维的关键点：

（1）不同状态之间的转换：为每个状态设置一个M和m，对于每一个后进入的点判断是否可以继续维持前一状态，如果可以就count++

（2）要意识到新状态的开始点并不一定是旧状态的结束点：可能有一点使得原来的状态不能够持续下去而结束了，但是并不意味着这一点就是新状态的开始点，在维持旧状态的过程中可能就出现了新状态的开始点，这点主要到了以后只要根据不同的情况去找那个开始点就OK了

---

#include <iostream>
using namespace std;

int m,M,n;
int num[100007];

bool ok(int t)
{
    if(m==M) {
        if(t==m)
            return true;
        else if(t == m-1||t == m+1)
            return true;
        else 
            return false;
    }
    else {
        if(t==M||t==m)
            return true;
        else 
            return false;
    }
}

int main()
{
    while(cin>>n)
    {
        cin>>num[1];
        m = num[1];
        M = num[1];
        int count = 1;
        int ans = 0;
        for(int i = 2;i <= n;i++)
        {
            cin>>num[i];
            if(ok(num[i]))
            {
                if(m==M && num[i] == m-1) 
                    m = num[i];
                else if(m==M && num[i] == m+1)
                    M = num[i];
                count++;
            }
            else {
                if(m == M) {
                    M = m = num[i];
                    count = 1;
                }
                else {
                    int pos;
                    if(num[i] == num[i-1]+1) {
                        M = num[i];
                        m = num[i-1];
                        for(int j = i-1;j >= 1;j--)
                        {
                            if(num[j] != num[i-1])
                                break;
                            pos = j;
                        }
                        count = i-pos+1;//此时的count应该=当前的坐标-s段开始的坐标 
                    }
                    else if(num[i] == num[i-1]-1) {
                        M = num[i-1];
                        m = num[i];
                        for(int j = i-1;j >= 1;j--)
                        {
                            if(num[j] != num[i-1])
                                break;
                            pos = j;
                        }
                        count = i-pos+1;//同上 
                    }
                    else {
                        M = m = num[i];
                        count = 1;
                    }
                }
            }
            ans = max(ans,count);
        }    
        cout<<ans<<endl;
    }
    return 0;
}