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.Scanner;

// https://codeforces.com/problemset/problem/1272/D
public class RemoveOneElement {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int arr[] = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		int dp[] = new int[n];
		dp[0] = 1;
		int ans = 1, prev = 0;
		for (int i = 1; i < n; i++) {
			if (arr[i] > arr[i - 1]) {
				dp[i] = dp[i-1] + 1;
			}
			else {
				ans = Math.max(ans, i - prev);
				dp[i] = 1;
				prev = i;
			}
		}
		if (dp[n-1] != 1) {
			ans = Math.max(ans, dp[n-1]);
		}
//		for (int i = 0; i < n; i++) {
//			System.out.print(dp[i]+ " ");
//		}
		int dp1[] = new int[n];
		dp1[n-1] = 1;
		for (int i = n - 2; i >= 0; i--) {
			if (arr[i] < arr[i + 1]) {
				dp1[i] = dp1[i+1] + 1;
			}
			else {
				dp1[i] = 1;
			}
		}
		for (int i = 1; i < n - 1; i++) {
			if (arr[i-1] < arr[i+1])
				ans = Math.max(ans, dp[i-1] + dp1[i + 1]);
		}
		System.out.println(ans);

	}

}
