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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.stream.IntStream;
import java.util.Arrays;
import java.util.Scanner;
import java.util.OptionalInt;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
            }
            int[] l = new int[n];
            Arrays.fill(l, 1);
            for (int i = 1; i < n; i++) {
                l[i] = a[i] > a[i - 1] ? l[i - 1] + 1 : 1;
            }
            int[] r = new int[n];
            Arrays.fill(r, 1);
            for (int i = n - 2; i >= 0; i--) {
                r[i] = a[i + 1] > a[i] ? r[i + 1] + 1 : 1;
            }
            int ans = Arrays.stream(l).max().getAsInt();
            for (int i = 1; i < n - 1; i++) {
                ans = a[i + 1] > a[i - 1] ? (Math.max(ans, l[i - 1] + r[i + 1])) : ans;
            }
            out.print(ans);
        }

    }
}

