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 Arange {

     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();
        }
        
        System.out.println(solve(a,n));
  
     }
     static int solve(int a[],int n){
         int left[]=new int [n];
         int right[]=new int [n];
         
         int ans=1;
         
         Arrays.fill(left,1);
         Arrays.fill(right,1);
         
         for(int i=1;i<n;i++){
             if(a[i]>a[i-1]){
                 left[i]=left[i-1]+1;
                 ans=Math.max(ans,left[i]);
             }
         }
         
          
         
          for(int i=n-2;i>=0;i--){
              if(a[i]<a[i+1]){
                  right[i]=right[i+1]+1;
                  ans=Math.max(ans,right[i]);
              }
          }
          
          for(int i=1;i<n-1;i++){
              if(a[i+1]>a[i-1])
              ans=Math.max(ans,left[i-1]+right[i+1]);
          }
          
          
          return ans;
     }
     
}
     