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.*;
import java.math.*;
public class Main
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=1;
		int n=sc.nextInt();
		int[] arr=new int[n+2];
		arr[n+1]=arr[n]=-1;
		for(int i=0;i<n;i++)
		arr[i]=sc.nextInt();
		int k=Integer.MAX_VALUE;
		int[][] dp=new int[n+1][1];
		dp[n-1][0]=1;
		dp[n][0]=0;
		int ans=1;
		for(int i=n-2;i>=0;i--)
		{
		    if(arr[i+1]>arr[i])
		    dp[i][0]=dp[i+1][0]+1;
		    else
		    dp[i][0]=1;
		    ans=Math.max(ans,dp[i][0]);
		    if((dp[i][0]!=1)&&(arr[i+dp[i][0]-2]<arr[i+dp[i][0]]))
		    ans=Math.max(ans,dp[i][0]-1+dp[i+dp[i][0]][0]);
		    if((dp[i][0]!=1)&&(arr[i+dp[i][0]-1]<arr[i+dp[i][0]+1]))
		    ans=Math.max(ans,dp[i][0]+dp[i+dp[i][0]+1][0]);
		    
		}
		System.out.println(ans);
	}
}
