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 D {

	private static Scanner in;
	private static int n;
	private static int[] a;
	
	public static void main(String[] args) {
		in = new Scanner(System.in);
		n = in.nextInt();
		a = new int[n];
		for(int i=0; i<n; i++) a[i] = in.nextInt();
		
		long[] left = new long[n];
		long[] right = new long[n];
		
		left[0] = 1;
		right[n-1] = 1;
		
		for(int i=1; i<n; i++) {
			if (a[i]>a[i-1]) {
				left[i] = left[i-1]+1;
			} else {
				left[i] = 1;
			}
		}
		
		for(int i=n-2; i>=0; i--) {
			if (a[i]<a[i+1]) {
				right[i] = right[i+1]+1;
			} else {
				right[i] = 1;
			}
		}
		
		long ans = 1;
		for(long x: left) ans = Math.max(ans, x);
		for(long y: right) ans = Math.max(ans, y);
		
		
		for(int i=1; i<n-1; i++) {
			if (a[i-1]<a[i+1]) {
				ans = Math.max(ans, left[i-1]+right[i+1]);
			}
		}
		System.out.println(ans);
	}

}
