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.*;
import java.lang.*;
public class pD{
public static void main(String[] args) throws FileNotFoundException, IOException{
       Scanner s=new Scanner(System.in);
      int n=s.nextInt();
        int[] arr=new int[n];
      for(int  i=0;i<n;i++){arr[i]=s.nextInt();}
           int[] dp1=new int[n];
      int[] dp2=new int[n];
      dp1[0]=1;
      for(int i=1;i<n;i++){
        if(arr[i]>arr[i-1]){
          dp1[i]=dp1[i-1]+1;
        }
        else{dp1[i]=1;}
      }

      dp2[0]=1;
       if(arr[0]<arr[1]){dp2[1]=2;}
       else{dp2[1]=1;}
      for(int i=2;i<n;i++){
        dp2[i]=1;
        if(arr[i]>arr[i-1]){
          dp2[i]=dp2[i-1]+1;
        }
// Fcuk it!! mannn... See what we missed during the contest
// when arr[i]>arr[i-1] then we have two choice
        // either we can take arr[i-1] into account or not
        // but we have consider only the first case i.e. we
        // have taken this into account with surety which is not correct
        
          if(arr[i-2]<arr[i]){
            dp2[i]=Math.max(dp1[i-2]+1,dp2[i]);
          }
          
        
      }
      int res=0;
      for(int i=0;i<n;i++){
        int k=Math.max(dp1[i],dp2[i]);
        res=Math.max(k,res);
      }
      System.out.println(res);
        
      }
        }
    
