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.
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1272D extends PrintWriter {
	CF1272D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1272D o = new CF1272D(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		int[] pp = new int[n];
		for (int i = 1; i < n; i++)
			if (aa[i - 1] < aa[i])
				pp[i] = pp[i - 1] + 1;
		int[] qq = new int[n];
		for (int i = n - 2; i >= 0; i--)
			if (aa[i] < aa[i + 1])
				qq[i] = qq[i + 1] + 1;
		int ans = 0;
		for (int i = 0; i < n; i++)
			ans = Math.max(ans, pp[i] + 1);
		for (int i = 1; i + 1 < n; i++)
			if (aa[i - 1] < aa[i + 1])
				ans = Math.max(ans, pp[i - 1] + 1 + qq[i + 1] + 1);
		println(ans);
	}
}
