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.
// package day6;

import java.util.*;

public class Codeforces1272D {

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

}
