Codeforces Round #FF (Div. 2)：C. DZY Loves Sequences

C. DZY Loves Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

DZY has a sequence a, consisting of n integers.

We'll call a sequence a_i, a_i + 1, ..., a_j (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment.

Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.

You only need to output the length of the subsegment you find.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵). The next line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

In a single line print the answer to the problem — the maximum length of the required subsegment.

Sample test(s)

Input

6
7 2 3 1 5 6

Output

5

Note

You can choose subsegment a₂, a₃, a₄, a₅, a₆ and change its 3rd element (that is a₄) to 4.

题意就是， 你能够改变字符串中的一个字符。 就出其最长的连续字串

如案列， 7 2 3 1 5 6 —————7 2 3  4 5 6

输出即为5.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>

using namespace std;

#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, m) for(int i=1; i<=m; i++)
#define f3(i, n) for(int i=n; i>=1; i--)
#define f4(i, n) for(int i=1; i<=n; i++)
#define f5(i, n) for(int i=2; i<=n; i++)
#define M 1005

const int INF = 0x3f3f3f3f;
int n, a[100005], b[100005];

int main()
{
    cin>>n;
    f4(i, n)
    cin>>a[i];
    b[1]=1;
    f5(i, n)
    {
        b[i]=1;
        if (a[i]>a[i-1])
            b[i]=b[i-1]+1;
    }
    int ans=-INF;
    f3(i, n)
    {
        if (b[i]==n)
            ans=max(b[i], ans);
        else
            ans=max(ans, b[i]+1);
        if (a[i-b[i]+1]-1>a[i-b[i]-1])
            ans=max(ans,b[i]+b[i-b[i]]);
        if (a[i-b[i]+2]-1>a[i-b[i]])
            ans=max(ans,b[i]+b[i-b[i]]);
    }
    cout<<ans<<endl;
    return 0;
}