You are given an array a consisting of n integers.

You can remove at most one element from this array. Thus, the final length of the array is n-1 or n.

Your task is to calculate the maximum possible length of the strictly increasing contiguous subarray of the remaining array.

Recall that the contiguous subarray a with indices from l to r is a[l ... r] = a_l, a_{l + 1}, ..., a_r. The subarray a[l ... r] is called strictly increasing if a_l < a_{l+1} < ... < a_r.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the maximum possible length of the strictly increasing contiguous subarray of the array a after removing at most one element.

Examples

Input


5
1 2 5 3 4


Output


4


Input


2
1 2


Output


2


Input


7
6 5 4 3 2 4 3


Output


2

Note

In the first example, you can delete a_3=5. Then the resulting array will be equal to [1, 2, 3, 4] and the length of its largest increasing subarray will be equal to 4.
import java.util.*;
public class MyClass {
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int a[]=new int[n+1];
        for(int i=1;i<=n;i++)
        {
            a[i]=sc.nextInt();
        }
        int l[]=new int[n+1];
        int r[]=new int[n+1];
        Arrays.fill(l,1);Arrays.fill(r,1);
        l[0]=0;
        int max=1;
        for(int i=2;i<=n;i++)
        {
            if(a[i]>a[i-1])
            {
                l[i]=l[i-1]+1;
                max=Math.max(max,l[i]);
            }
        }
        for(int i=n-1;i>=1;i--)
        {
            if(a[i]<a[i+1])
            {
                r[i]=r[i+1]+1;
                max=Math.max(max,r[i]);
            }
        }
        for(int i=1;i<=n-2;i++)
        {
            if(a[i]<a[i+2])
            {
                max=Math.max(max,l[i]+r[i+2]);
            }
        }
        System.out.println(max);
    }
}