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.
/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

public class GFG {
	public static void main (String[] args) {
	    
	    Scanner sc = new Scanner(System.in);
	    
	    int n = sc.nextInt();
	    
	    int a[] = new int[n];
	    for(int i=0;i<n;i++)
	    a[i] = sc.nextInt();
	 
	  int l[] = new int[n];
	  Arrays.fill(l,1);
	  
	  int r[] = new int[n];
	  Arrays.fill(r,1);
	  
	  int ans=1;
	  
	  
	  for(int i=1;i<n;i++)
	  {
	      if(a[i] > a[i-1]){
	        l[i] = l[i-1]+1;
	        ans = Math.max(ans,l[i]);
	      }
	  }
	  
	  for(int i=n-2;i>=0;i--){
	      if(a[i] < a[i+1]){
	      r[i] = r[i+1]+1;
	      ans = Math.max(ans,r[i]);
	      }
	  }
	 
     for(int i=0;i<n-2;i++){
         if(a[i]<a[i+2]){
             
             ans=Math.max(ans,l[i]+r[i+2]);
         }
    }
    
    System.out.println(ans);
	    
	}
   
}