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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

public class O {
	public static void main(String[] args) throws IOException {
		StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		st.nextToken();
		int n = (int) st.nval;
		int[] a = new int[n];
		int[] b = new int[n];
		int[] c = new int[n];
		for (int i = 0; i < n; i++) {
			st.nextToken();
			a[i] = (int) st.nval;
		}
		b[0] = 1;
		for (int i = 1; i < n; i++)
			if (a[i] > a[i - 1])
				b[i] = b[i - 1] + 1;
			else
				b[i] = 1;
		c[n - 1] = 1;
		for (int i = n - 2; i >= 0; i--)
			if (a[i] < a[i + 1])
				c[i] = c[i + 1] + 1;
			else
				c[i] = 1;
		int max = 0;
		for (int i = 0; i < n; i++)
			max = Math.max(max, b[i]);
		for (int i = 1; i < n - 1; i++)
			if (a[i - 1] < a[i + 1])
				max = Math.max(max, b[i - 1] + c[i + 1]);

		System.out.println(max);
	}
}

 	   					 	 					   			  		 	