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;
import java.util.*;

public class gg {
	
	

	public static void main(String[] args) {
	   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 [] pre = new int[n+1];
	   int[] suff = new int[n+1];
	   pre[0] = 0;
	   pre[1] = 1;
	   int max2 = 0;
	   for(int i=1; i<n; i++) {
		   if(arr[i] > arr[i-1]) {
			   pre[i+1] = pre[i]+1;
			   max2 = Math.max(pre[i+1], max2);
		   }
		   else {
			   pre[i+1] = 1;
		   }
	   }
	   suff[n] = 0;
	   suff[n-1] = 1;
	   for(int i =n-2; i>=0; i--) {
		   if(arr[i] < arr[i+1]) {
			   suff[i] = suff[i+1]+1;
		   }
		   else {
			   suff[i] = 1;
		   }
	   }
	   
	   int max = 0;
	   for(int i =0; i<n; i++) {
		   int sum = pre[i] + suff[i+1];
		   if(i==0 || i==n-1) {
			   max = Math.max(max,  sum);
		   }
		   else {
			   if(arr[i-1] < arr[i+1]) {
				   max = Math.max(max, sum);
			   }
		   }
	   }
	   System.out.println(Math.max(max2, max));
	}

}
