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.
// template halnix16


import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {


    static BufferedReader br;

    public static void main(String[] args) throws Exception {
        // write your code here
        br = new BufferedReader(new InputStreamReader(System.in));
        long n = in();
        long arr[] = arr(n);
        long dp[][] = new long[(int)n][2];

        dp[0][0]=1;
        for(int k=1;k<n;k++){

            if(arr[k]>arr[k-1]){
                dp[k][0]= dp[k-1][0]+1;
            }else{
                dp[k][0] = 1;
            }


        }
        dp[(int)n-1][1] = 1;

        for(int k =(int)n-2;k>1;k--){

            if(arr[k]>arr[k+1]){
                dp[k][1] =1 ;
            }else
                dp[k][1]= dp[k+1][1]+1;

        }
        long ans=0;


        for(int i =0;i<n;i++){
            ans = Math.max(ans,dp[i][0]);
            if(i>=n-2)
                continue;
                if(arr[i+2]>arr[i]) {
                    ans = Math.max(ans, dp[i][0] + dp[i + 2][1]);
                }

            


        }
        System.out.println(ans);



    }



    static long[] arr(long n) throws Exception {

        String read[] = sarrin();
        long arr[] = new long[(int) n];
        for (int k = 0; k < n; k++)
            arr[k] = in(read[k]);
        return arr;

    }

    static long[][] arr(long n, long m) throws Exception {


        long arr[][] = new long[(int) n][(int) m];
        for (int k = 0; k < n; k++) {
            String read[] = sarrin();
            for (int j = 0; j < m; j++)
                arr[k][j] = in(read[k]);
        }
        return arr;

    }

    static long in() throws Exception {
        return Long.parseLong(br.readLine());


    }

    static long in(String num) throws Exception {
        return Long.parseLong(num);


    }


    static String sin() throws Exception {

        return br.readLine();
    }

    static String[] sarrin() throws Exception {

        return br.readLine().split(" ");

    }

    static String[] regexSplit() throws Exception {
        return br.readLine().split(" ?(?<!\\G)((?<=[^\\p{Punct}])(?=\\p{Punct})|\\b) ?");
    }


}





