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.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {
	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 < a.length; ++i) {
			a[i] = sc.nextInt();
		}
		System.out.println(solve(a));

		sc.close();
	}

	static int solve(int[] a) {
		int n = a.length;

		int[] leftLengths = new int[n];
		leftLengths[0] = 1;
		for (int i = 1; i < leftLengths.length; ++i) {
			if (a[i] > a[i - 1]) {
				leftLengths[i] = leftLengths[i - 1] + 1;
			} else {
				leftLengths[i] = 1;
			}
		}

		int[] rightLengths = new int[n];
		rightLengths[rightLengths.length - 1] = 1;
		for (int i = rightLengths.length - 2; i >= 0; --i) {
			if (a[i] < a[i + 1]) {
				rightLengths[i] = rightLengths[i + 1] + 1;
			} else {
				rightLengths[i] = 1;
			}
		}

		return Math.max(Arrays.stream(leftLengths).max().getAsInt(),
				IntStream.rangeClosed(1, n - 2).filter(i -> a[i - 1] < a[i + 1])
						.map(i -> leftLengths[i - 1] + rightLengths[i + 1]).max().orElse(Integer.MIN_VALUE));
	}
}
