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.InputStreamReader;



public class Main {
	public static void main(String[] args) {
		try(BufferedReader br = new BufferedReader(new InputStreamReader(System.in))){
			int n = Integer.parseInt(br.readLine());
			int numbers[] = new int[n];
			String s[] = br.readLine().split(" ");
			for (int i = 0 ; i < n; ++i) {
				numbers[i] = Integer.parseInt(s[i]);
			}
			int l[] = new int[n];
			int r[] = new int[n];
			l[0] = 1;
			r[n-1] = 1;
			int ans = 0;
			for(int i = 1; i < n; ++i) {
				if(numbers[i]>numbers[i-1]) {
					l[i] = l[i-1]+1;
					ans = Math.max(ans, l[i]);
				}
					
				else
					l[i] = 1;
			}
			for(int i = n-1 ; i > 0; --i) {
				if(numbers[i-1]<numbers[i])
					r[i-1] = r[i] + 1;
				else
					r[i-1] = 1;
				
			}
			
			for(int i = 1; i <= n-2; ++i) {
				if(numbers[i+1]-numbers[i-1] > 0)
					ans = Math.max(ans, l[i-1]+r[i+1]);
			}
			System.out.println(Math.max(ans, r[0]));
		}
		catch(Exception e) {
			e.printStackTrace();
		}
	}

}
