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.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int size = scanner.nextInt();
        long[] a = new long[size];
        Integer[] resultA = new Integer[size];
        Integer[] resultB = new Integer[size];
        for (int i=0; i<size;i++) {
            a[i] = scanner.nextLong();
            resultA[i] = 1;
            resultB[i] = 1;
        }
        for (int i=size-2;i>=0;i--){
            if(a[i]<a[i+1]) {
                resultA[i] = resultA[i + 1] + 1;
                resultB[i] = resultB[i + 1] + 1;
            }
            if(i+2<size && a[i]<a[i+2])
                resultB[i]= Math.max(resultB[i],resultA[i+2]+1);
        }
        int maxA = Collections.max(Arrays.asList(resultA));
        int maxB = Collections.max(Arrays.asList(resultB));
        System.out.println(Math.max(maxA,maxB));
    }
}

  	 			  		 	   	 	 	 	    			