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.ArrayList;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        ArrayList<Integer> values = new ArrayList<>();
        for(int i=0; i<n; i++) {
            values.add(sc.nextInt());
        }
        int[][] best = new int[n][2];
        for(int i=0; i<n; i++) {
            best[i][0] = 1;
            best[i][1] = 1;
        }
        int max = 0;
        for(int i=1; i<n; i++) {
            if(values.get(i) > values.get(i-1)) best[i][0] = best[i-1][0]+1;
            if(values.get(i) > values.get(i-1)) best[i][1] = best[i-1][1]+1;
            if(i-2 >= 0 && values.get(i) > values.get(i-2)) best[i][1] = Math.max(best[i][1], best[i-2][0] + 1);
            max = Math.max(best[i][0], max);
            max = Math.max(best[i][1], max);
        }
        System.out.println(max);
    }

}
