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.*;
public class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int N = scan.nextInt();
        ArrayList<Integer> Arr = new ArrayList<>();
        for (int i=0;i<N;i++){
            Arr.add(scan.nextInt());
        }
        int[] arr = new int[N];
        Arrays.fill(arr, 1);
        int[] arr2 = new int[N];
        Arrays.fill(arr2, 1);
        int max = 1;
        for (int i=1;i<N;i++){
            if (Arr.get(i-1)<Arr.get(i)){
                arr[i] = Math.max(arr[i], arr[i-1]+1);
                max = Math.max(max, arr[i]);
            }
        }
        for (int i=N-2;i>=1;i--){
            if (Arr.get(i+1)>Arr.get(i)){
                arr2[i] = Math.max(arr2[i], arr2[i+1]+1);
                max = Math.max(max, arr2[i]);
            }
        }
        for (int i=1;i<N-1;i++){
            if (Arr.get(i-1)<Arr.get(i+1)){
                max = Math.max(max, arr[i-1]+arr2[i+1]);
            }
        }
        System.out.println(max);
    }
}