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.io.*;
public class Test
{
    static int a[];
    static long dp[][];
    // static int pref
    static int mod = 1000000007;
    
    public static void main(String[] args) throws Exception {   
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String s[] = in.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        
        s = in.readLine().split(" ");
        a = new int[n];
        for(int i = 0; i < n; i++){
            a[i] = Integer.parseInt(s[i]);
        }
        int endat[] = new int[n];
        endat[0] = 1;
        int max = 1;
        for(int i = 1; i < n; i++){
            if(a[i-1] < a[i])
                endat[i] = endat[i-1]+1;
            else{
                endat[i] = 1;
            }
            max = Math.max(max,endat[i]);
        }
        int startat[] = new int[n];
        startat[n-1] = 1;
        for(int i = n-2; i >= 0; i--){
            if(a[i+1] > a[i])
                startat[i] = startat[i+1]+1;
            else{
                startat[i] = 1;
            }
        }
        
        for(int i = 1; i < n-1;i++){
            if(a[i-1] < a[i+1]){
                max = Math.max(max,startat[i+1]+endat[i-1]);
            }
        }
        System.out.println(max);
 
    } 
}
