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.Scanner;

public class Main {

	public static void main(String[] args) 
	{
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int a[] = new int[n];
		for(int i=0;i<n;i++)a[i]=scan.nextInt();
		System.out.println(maxIncSubarr(a, n));

	}
	static int maxIncSubarr(int a[], int n)  
    {  
        // Create two arrays pre and pos  
        int pre[] = new int[n] ;  
        int pos[] = new int[n] ;  
        pre[0] = 1;  
        pos[n - 1] = 1;  
        int l = 0;  
      
        // Find out the contribution of the current  
        // element in array[0, i] and update pre[i]  
        for (int i = 1; i < n; i++) 
        {  
            if (a[i] > a[i - 1])  
                pre[i] = pre[i - 1] + 1;  
            else
                pre[i] = 1;  
        }  
      
        // Find out the contribution of the current  
        // element in array[N - 1, i] and update pos[i]  
        l = 1;  
        for (int i = n - 2; i >= 0; i--) 
        {  
            if (a[i] < a[i + 1])  
                pos[i] = pos[i + 1] + 1;  
            else
                pos[i] = 1;  
        }  
      
        // Calculate the maximum length of the  
        // stricly increasing subarray without  
        // removing any element  
        int ans = 0;  
        l = 1;  
        for (int i = 1; i < n; i++) 
        {  
            if (a[i] > a[i - 1])  
                l++;  
            else
                l = 1;  
            ans = Math.max(ans, l);  
        }  
      
        // Calculate the maximum length of the  
        // strictly increasing subarray after  
        // removing the current element  
        for (int i = 1; i <= n - 2; i++)  
        {  
            if (a[i - 1] < a[i + 1])  
                ans = Math.max(pre[i - 1] +  
                                pos[i + 1], ans);  
        }  
        return ans;  
    }  
        
} 
