DZY Loves Sequences

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 446A

Description

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 Input

Input

6
7 2 3 1 5 6

Output

5

Hint

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

代码有毒！ 调了好久好久。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int i,j;
    int b[100005],c[100005];
    int n,a[100005],ma;
    while(scanf("%d",&n)!=EOF)
    {
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        ma=0;
        a[0]=0,a[n+1]=0,b[0]=0,b[n+1]=0,c[0]=0,c[n+1]=0;
        for(i=1;i<=n;i++)
        {
            if(a[i]>a[i-1])
                b[i]=b[i-1]+1;
            else
                b[i]=1;
            if(b[i]>ma)
                ma=b[i];
        }
        for(i=n;i>=1;i--)
        {
            if(a[i]<a[i+1])
                c[i]=c[i+1]+1;
            else
                c[i]=1;
            if(c[i]>ma)
                ma=c[i];
        }
        for(i=1;i<=n;i++)
        {
            if(a[i+1]-a[i-1]>=2)
            {
                ma=max(ma,b[i-1]+c[i+1]+1);
            }
            else
            {
                ma=max(ma,max(b[i-1]+1,c[i+1]+1));
            }
        }
        printf("%d\n",ma);
    }
    return 0;
}