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 RemoveElement {
 
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int[] a = new int[n];
		
		for(int i=0;i<n;i++)
			a[i]=input.nextInt();
		
		int[] start = new int[n];
		int[] end = new int[n];
		
		end[0]=1;
		start[n-1]=1;
		
		int max = 1;
		for(int i=1;i<n;i++) {
			if(a[i]>a[i-1])
				end[i]+=end[i-1]+1;
			else
				end[i]=1;
			max =  Math.max(max, end[i]);
		}
		
		for(int  i=n-2;i>=0;i--) {
			if(a[i]<a[i+1])
				start[i]+=start[i+1]+1;
			else
				start[i]=1;
			max = Math.max(start[i],max);
		}
		
		for(int i=1;i<n-1;i++) {
			if(a[i+1]>a[i-1])
				max = Math.max(max, end[i-1]+start[i+1]);
		}
		
		System.out.println(max);
	}
 
}