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 RemoveOneElement {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
            arr[i] = s.nextInt();

        int[] ml = new int[n];
        int ret = 1;
        int i = 1, cnt = 1, j = 0;
        while (i <= n) {
            if (i < n && arr[i-1] < arr[i]) {
                cnt++;
                ret = Math.max(ret, cnt);
            } else {
                while (j < i)
                    ml[j++] = cnt--;

                cnt = 1;
            }
            i++;
        }

        //System.out.println(Arrays.toString(ml));

        for (i = 0; i < arr.length - 2; i++) {
            if ((i + ml[i] + 1) < arr.length && arr[i + ml[i] + 1] > arr[i + ml[i] - 1])
                ret = Math.max(ret, ml[i] + ml[i+ ml[i]+1]);
            else if (i + ml[i] < arr.length && i + ml[i] - 2 >= 0 && arr[i + ml[i]] > arr[i + ml[i] - 2])
                ret = Math.max(ret, ml[i] - 1 + ml[i + ml[i]]);
        }

        System.out.println(ret);

    }
}