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.io.*;
import java.util.*;

public class D {

    public static void main(String args[]) {
        Scanner scn = new Scanner(System.in);
        int n = scn.nextInt();
        int a[] = new int[n];
        for (int i = 0; i < n; ++i) {
            a[i] = scn.nextInt();
        }
        int max = 1;
        int fwd[] = new int[n + 1];
        int bck[] = new int[n + 1];
        fwd[0] = 0;
        fwd[1] = 1;
        bck[n] = 0;
        bck[n - 1] = 1;
        int count = 1;
        for (int i = 1; i < n; ++i) {
            if (a[i] > a[i - 1]) {
                count++;
            } else {
                count = 1;
            }
            fwd[i + 1] = count;
        }
        count = 1;
        for (int i = n - 2; i > 0; --i) {
            if (a[i] < a[i + 1]) {
                count++;
            } else {
                count = 1;
            }
            bck[i] = count;
        }
        boolean nobreak = true;
        for (int i = 0; i <= n - 1; ++i) {
            if (fwd[i] > fwd[i + 1]) {
                if (i - 2 >= 0 && a[i - 2] < a[i]) {
                    max = Math.max(fwd[i] + bck[i] - 1, max);
                    nobreak = false;
                }
                if (i - 1 >= 0 && i + 1 < n && a[i - 1] < a[i + 1]) {
                    max = Math.max(fwd[i] + bck[i] - 1, max);
                    nobreak = false;
                }
                if (i == n - 1) {
                    max = Math.max(fwd[i] + bck[i] - 1, max);
                }
            }
        }
        for (int i = 0; i <= n; ++i) {
            max = Math.max(max, fwd[i]);
        }
        if (nobreak)
            max = Math.max(fwd[n] + bck[n], max);
        System.out.println(max);
    }
}