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.*;
import java.io.*;
public class RemoveOneElement{
	static int compute(int A[], int n)
	{
		return n;
	}
	public static void main(String[] args) throws IOException {
	 BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); 
		Scanner sc = new Scanner(System.in);
		int t=1;
		while(t-- >0)
		{
			// HashMap<Integer, Integer> hm=new HashMap<Integer, Inetger>();
			// ArrayList<Integer> al=new ArrayList<Integer>();
			int n=sc.nextInt();
			int A[]=new int[n+1]; 			
			int max=1;
			int Con[]=new int[n+1];
            int Rem[]=new int[n+1];
            A[0] = Integer.MIN_VALUE;
            Con[0] = 0;
            Con[1] = 1;
            Rem[0] = 0;
            Rem[1] = 1;
			for(int i=1;i<=n;i++)
                A[i]=sc.nextInt();
            for(int i=2;i<=n;i++) {
                if(A[i]>A[i-1]) {
                    Con[i] = Con[i-1] + 1;
                    Rem[i] = Rem[i-1] + 1;
                    if(A[i] > A[i-2])
                        Rem[i] = Math.max(Rem[i-1], Con[i-2]) + 1;
                }
                else if(A[i] > A[i-2]) {
                    Con[i] = 1;
                    Rem[i] = Con[i-2] + 1;
                }
                else {
                    Con[i] = 1;
                    Rem[i] = 1;
                }
                max=Math.max(max, Con[i]);
                max=Math.max(max, Rem[i]);
            }
			System.out.println(max);
		}
	}
}