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 RemoveOneElement {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);

		int n = input.nextInt();

		int[] a = new int[n + 1];
		for (int i = 1; i <= n; i++) {
			a[i] = input.nextInt();
		}

		input.close();

		int[] f = new int[n + 1];
		f[1] = 1;
		for (int i = 2; i <= n; i++) {
			if (a[i] > a[i - 1]) {
				f[i] = f[i - 1] + 1;
			} else {
				f[i] = 1;
			}
		}

		int[] s = new int[n + 1];
		s[n] = 1;
		for (int i = n - 1; i >= 1; i--) {
			if (a[i] < a[i + 1]) {
				s[i] = s[i + 1] + 1;
			} else {
				s[i] = 1;
			}
		}

		int ans = 0;
		for (int i = 1; i <= n; i++) {
			ans = Math.max(ans, f[i]);
		}

		for (int i = 2; i < n; i++) {
			if (a[i - 1] < a[i + 1]) {
				ans = Math.max(ans, f[i - 1] + s[i + 1]);
			}
		}

		System.out.println(ans);
	}
}
