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 wha {
	public static void main(String args[]) {
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int arr[] = new int[n];
		for(int i=0;i<n;i++) {
			arr[i] = input.nextInt();
		}
		System.out.println(sol(arr,arr.length));

	}
	static int sol(int arr[],int n) {
		int l[] = new int[n];
		int r[] = new int[n];
		for(int i=0;i<n;i++) {
			l[i] = r[i] = 1;
		} 
		for(int i=1;i<n;i++) {
			if(arr[i]>arr[i-1]) {
				l[i] = l[i-1]+1;
			}
		}
		int max = 1;
		for(int i=n-2;i>=0;i--) {
			if(arr[i+1]>arr[i]) {
				r[i] = r[i+1]+1;
				max = Math.max(max, r[i]);
			}
		}
		for(int i=1;i<n-1;i++) {
			if(arr[i+1]>arr[i-1]) {
				max = Math.max(max,l[i-1]+r[i+1]);
			}
		}
		return max;
	}
}
