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 solution
{
   public static void main(String args[])
   {
      Scanner sc=new Scanner(System.in);
      String a=sc.nextLine();
      
      String c[]=sc.nextLine().split(" ");
      int d[]=new int[c.length];
      for(int i=0;i<d.length;i++)
      {
         d[i]=Integer.parseInt(c[i]);
         
      }
      int dp[][]=new int [c.length][2];
      dp[0][0]=1;
      dp[0][1]=1;
      int max=1;
      for(int i=1;i<d.length;i++)
      {
         if(i==1)
         {
            dp[i][0]=1;
            dp[i][1]=1;
            if(d[i]>d[i-1])
            {
               dp[i][0]=2;
            }
         }
         else
         {
            dp[i][0]=1;
            dp[i][1]=1;
            int m=0;
            if(d[i]>d[i-1])
            {
               dp[i][0]+=dp[i-1][0];
               m=dp[i-1][1];
            }
            if(d[i]>d[i-2])
            {
               m=Math.max(m,dp[i-2][0]);
            }
            dp[i][1]+=m;
            
         }
         max=Math.max(max,Math.max(dp[i][1],dp[i][0]));
      }
      System.out.println(max);
   }
}