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.*;

public class D {
	static Scanner sc = new Scanner(System.in);

	static void test() {
		int n = sc.nextInt();
		int[] a = new int[n];

		for (int i = 0; i < n; i++)
			a[i] = sc.nextInt();

		int[] l = new int[n];
		int[] r = new int[n];

		int len = 0;
		int prev = a[0];
		for (int i = 0; i < n; i++) {
			if (a[i] <= prev) {
				len = 1;
			} else {
				len++;
			}
			l[i] = len;
			prev = a[i];
		}

		len = 0;
		int next = a[n - 1];
		for (int i = n - 1; i >= 0; i--) {
			if (a[i] >= next)
				len = 1;
			else
				len++;
			next = a[i];
			r[i] = len;
		}

		/*for (int i = 0; i < n; i++)
			System.out.print(l[i] + " ");
		System.out.println();
		for (int i = 0; i < n; i++)
			System.out.print(r[i] + " ");
		System.out.println();
		*/
		int max = 0;
		for (int i = 0; i < n; i++) {
			max = Math.max(l[i], max);
		}
		for (int i = 1; i < n - 1; i++) {
			if (a[i - 1] < a[i + 1]) {
				// if (l[i - 1] + r[i + 1] > max)
					// System.out.println("i == " + i);
				max = Math.max(l[i - 1] + r[i + 1], max);
			}
		}

		max = Math.max(l[n - 1], max);
		max = Math.max(l[n - 2], max);
		max = Math.max(r[1], max);

		System.out.println(max);
	}

	public static void main(String[] args) {
		test();
	}
}
