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.Arrays;
import java.util.Scanner;

public class D1272 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
       int n = scanner.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
        }
        process(a, n);
    }

    private static void process(int[] a, int n) {
        int[] orig = new int[n];
        int[] removed = new int[n];
        orig[0] = removed[0] = 1;
        removed[1] = 1;
        for (int i = 1; i < n; i++) {
            orig[i] = a[i-1] < a[i] ? orig[i-1] + 1 : 1;
        }

        for (int i = 2; i < n; i++) {
            int x = a[i-2] < a[i] ? orig[i - 2] + 1 : 1;
            int y = a[i-1] < a[i] ? removed[i - 1] + 1 : 1;
            removed[i] = Math.max(x, y);
        }
        int max = -1;
        for (int i = 0; i < n; i++) {
            max = Math.max(max, Math.max(orig[i], removed[i]));
        }
        System.out.println(max);
    }

}
