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.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
	try
	{
	BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	Integer n=Integer.parseInt(br.readLine());
	String [] starrr=br.readLine().trim().split("\\s+");
	long [] arr=new long[n];
	for(int i=0;i<n;i++)
	{
		arr[i]=Long.parseLong(starrr[i]);
	}
	System.out.println(solve(arr));
	}
	catch(Exception e)
	{
		e.printStackTrace();
	}
}
static int solve(long [] arr)
{
	int n=arr.length;
	Integer res=0;
	int [] dpl=new int[n];
	int [] dpr=new int[n];
	Arrays.fill(dpl, 1);
	Arrays.fill(dpr, 1);
	for(int i=1;i<n;i++)
	{
		if(arr[i]>arr[i-1])
		{
			dpl[i]=dpl[i-1]+1;
		}
		res=Math.max(res, dpl[i]);
	}
	for(int i=n-2;i>=0;i--)
	{
	     if(arr[i+1]>arr[i])
	     {
	    	 dpr[i]=dpr[i+1]+1;
	     }
	     res=Math.max(res, dpr[i]);
	}
	for(int i=0;i<n-2;i++)
	{
		//skipping one element now
		if(arr[i]<arr[i+2])
		{
			
			res=Math.max(res,dpl[i]+dpr[i+2] );
		}
	}
	return res;
}
}
