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 ai, ai + 1, ..., aj (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, ..., an (1 ≤ ai ≤ 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.

 

思路：预处理出包含第i个元素的左边最长连续递增长度 和 包含第i个元素的右边最长连续递增长度， 枚举可以任意改变的位置 pos， 然后判断该位置的左右两边是否可以连接以满足递增

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
const int N = 100005;
int a[N], lt[N], rt[N], n;
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    lt[1] = 1;
    for(int i = 2; i <= n; ++i) { //预处理i的左边且包含i的连续最长递增数列的长度
        if(a[i] > a[i - 1]) lt[i] = lt[i - 1] + 1;
        else lt[i] = 1;
    }

  //  for(int i = 1; i <= n; ++i) cout << lt[i] << ' ' ; cout  << endl;
    rt[n] = 1;
    for(int i = n - 1; i >= 1; --i) { ////预处理i的右边且包含i的连续最长递增数列的长度
        if(a[i] < a[i + 1]) rt[i] = rt[i + 1] + 1;
        else rt[i] = 1;
    }
  //  for(int i = 1; i <= n; ++i) cout << rt[i] << ' ' ; cout  << endl;
    int ans = max(rt[2] + 1, lt[n - 1] + 1);
    for(int i = 2; i < n; ++i) {//枚举可任意更换的位置
        if(a[i - 1] + 1 >= a[i + 1]) ans = max(ans, max(lt[i - 1], rt[i + 1]) + 1); // 若i的两边不能衔接， 取较大的长度
        else ans = max(ans, lt[i - 1] + rt[i + 1] + 1);//否则为答案为 两段和
    }
    printf("%d\n", ans);
}