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.*;
import java.io.*;
public class D1272{
    public static void main(String args[]) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        StringBuilder sb = new StringBuilder();
        String temp[] = br.readLine().split(" ");
        int arr[] = new int[n];
        PA(arr,temp);
        DP = new int[n][2];
        int max = -1;
        for(int i=0;i<n;i++){
            max = Math.max(max,solve(arr,i,0));
        }
        System.out.println(max);
    }
    static int DP[][];
    static int solve(int arr[],int pos,int delCount){
        if(pos == 0)
            return 1;
        if(DP[pos][delCount] !=0)
            return DP[pos][delCount];
        int max =1;
            if(arr[pos-1]<arr[pos]){
                max =Math.max(solve(arr, pos-1, delCount)+1,max);
            }
            if(delCount == 0 && (pos-2>=0) && arr[pos-2]<arr[pos]){
                max =Math.max(solve(arr, pos-2, delCount+1)+1,max);
            }
        return (DP[pos][delCount] = max);
    }

    static int PI(String s){
        return Integer.parseInt(s);
    }

    static void PA(int arr[],String temp[]){
        for(int i=0;i<arr.length;i++){
            arr[i] = PI(temp[i]);
        }
    }
}