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.
//package codeforces.div2;

import java.util.Scanner;

public class RemoveOneElement {

    static public void print(int arr[]) {
        int pref[] = new int[arr.length];

        int suff[] = new int[arr.length];

        int count = 1;
        pref[0] = count;
        for(int i = 1; i < arr.length; i++) {
            if(arr[i] > arr[i-1]) {
                count++;
            } else {
                count = 1;
            }
            pref[i] = count;
        }
        count = 1;
        suff[arr.length - 1] = count;

        for(int i = arr.length - 2; i>=0; i--) {
            if(arr[i]  < arr[i+1]) {
                count++;
            } else {
                count = 1;
            }
            suff[i] = count;
        }
        int maxCount = 1;

        for(int i = 0; i < arr.length ; i++) {
            if((i-1)>=0 && i+1 < arr.length && arr[i-1] < arr[i+1]) {
                maxCount = Math.max(pref[i - 1] + suff[i + 1], maxCount);
            } else {
                if((i-1)>=0) {
                    maxCount = Math.max(maxCount, pref[i-1]);
                } else {
                    if(i + 1 < arr.length) {
                        maxCount = Math.max(maxCount, suff[i+1]);
                    } else {

                    }
                }
            }
            if((i-1)>=0 && arr[i] > arr[i-1]) {
                maxCount = Math.max(maxCount, pref[i]);
            }
        }
        System.out.println(maxCount);
    }

    public static void main(String []args) {

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int i = 0;
        int arr[] = new int[n];
        while ( i < n) {
            arr[i++] = sc.nextInt();
        }
        print(arr);
    }
}
