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 IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        StringTokenizer st;
        int n = Integer.parseInt(br.readLine());
        int[] nums = new int[n];
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i<n; i++) {
            nums[i] = Integer.parseInt(st.nextToken());
        }
        int ans = 0;
        int[] l = new int[n];
        Arrays.fill(l,1);
        for (int i = 1; i<n; i++) {
            if (nums[i-1] < nums[i]) {
                l[i] = l[i-1]+1;
            }
            ans = Math.max(ans,l[i]);
        }
        int[] r = new int[n]; 
        Arrays.fill(r,1);
        for (int i = n-2; i>=0; i--) {
            if (nums[i+1] > nums[i]) {
                r[i] = r[i+1]+1;
            }
            ans = Math.max(ans,r[i]);
        }
        for (int i = 0; i<n-2; i++) {
            if (nums[i] < nums[i+2]) {
                ans = Math.max(ans,l[i]+r[i+2]);
            }
        }
        pw.println(ans);
        br.close();
        pw.close();
    }
}