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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.StringTokenizer;

public class D {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int l = Integer.parseInt(br.readLine());
        int[] arr = new int[l];
        StringTokenizer tokenizer = new StringTokenizer(br.readLine());
        for (int a = 0; a < l; a++)
            arr[a] = Integer.parseInt(tokenizer.nextToken());

        int[] dp = new int[l];
        int[] dpr = new int[l];
        dp[l - 1] = 1;
        dpr[0] = 1;
        int out = 1;
        ArrayList<Integer> list = new ArrayList<Integer>();
        for (int a = l - 2; a >= 0; a--) {
            if (arr[a] < arr[a + 1])
                out = Math.max(out, dp[a] = dp[a + 1] + 1);
            else {
                dp[a] = 1;
                list.add(a);
            }
        }
        for (int a = 1; a < l; a++) {
            if (arr[a] > arr[a - 1])
                out = Math.max(out, dpr[a] = dpr[a - 1] + 1);
            else {
                dpr[a] = 1;
                list.add(a);
            }
        }
        for(int a : list){
            if(a != 0 & a != l - 1){
                if(arr[a - 1] < arr[a + 1])
                    out = Math.max(out, dpr[a - 1] + dp[a + 1]);
            }
        }

        System.out.println(out);
    }
}
