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 Hello
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n;
        n=sc.nextInt();
        int[] v=new int[n];
        for(int i=0;i<n;i++)
            v[i]=sc.nextInt();
        int[] dp1=new int[200005],dp2=new int[200005];
        dp1[0]=1;
        for(int i=1;i<n;i++)
        {
              if(v[i]>v[i-1])
                  dp1[i]=dp1[i-1]+1;
              else
                dp1[i]=1;
        }
        dp2[n-1]=1;
        for(int i=n-2;i>=0;i--)
        {
            if(v[i]<v[i+1])
                dp2[i]=dp2[i+1]+1;
            else
                dp2[i]=1;
        }
        int ans=0;
        for(int i=1;i<n-1;i++)
        {
            if(v[i+1]>v[i-1])
                ans=Math.max(ans,dp1[i-1]+dp2[i+1]);
            else
                ans=Math.max(ans,Math.max(dp1[i-1],dp2[i+1]));
        }
        ans=Math.max(dp1[1],ans);
        ans=Math.max(dp1[2],ans);
        for(int i=0;i<n;i++)
        {
            ans=Math.max(ans,dp1[i]);
            ans=Math.max(ans,dp2[i]);
        }
        System.out.println(ans);
    }
}
