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