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.util.*; 
public class Main{
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
//		String[] cmd=br.readLine().split(" ");
		int cases=1;
		while(cases!=0)
		{
			cases--;
			int n=Integer.valueOf(br.readLine());
			int[] arr=new int[n];
			String[] cmd=br.readLine().split(" ");
			for(int i=0;i<n;i++)
			{
				arr[i]=Integer.valueOf(cmd[i]);
			}
			int[] nor=new int[n];
			int[] fin=new int[n];
			Arrays.fill(nor,1);
			Arrays.fill(fin,1);
			for(int i=1;i<n;i++)
			{
				if(arr[i]>arr[i-1])
				{
					nor[i]=nor[i-1]+1;
					fin[i]=Math.max(fin[i],fin[i-1]+1);
				}
				if(i-2>=0 && arr[i]>arr[i-2])
					fin[i]=Math.max(fin[i],nor[i-2]+1);
//				fin[i]=Math.max(fin[i],nor[i]);
			}
			int ans=0;
//			System.out.println(Arrays.toString(fin));
			for(int i=0;i<n;i++)
			{
				ans=Math.max(ans,fin[i]);
			}
			System.out.println(ans);
		}
	}

}
