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.*;
public class Abc {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int ar[] = new int[n];
		for(int i = 0; i < n; i++) {
			ar[i] = in.nextInt();
		}
		int dp[] = new int[n];
		dp[0] = 1;
		for(int i = 1; i < n ;i++) {
			if(ar[i] > ar[i - 1]) {
				dp[i] = dp[i - 1] + 1;
			} else {
				dp[i] = 1;
			}
		}
		int bp[] = new int[n];
		bp[n - 1] = 1;
		for(int i = n - 2; i >= 0; i--) {
			if(ar[i] < ar[i + 1]) {
				bp[i] = bp[i + 1] + 1;
			} else {
				bp[i] = 1;
			}
		}

		int maxVal = 0;
		for(int i = 0; i < n; i++) {
			maxVal = Math.max(maxVal, dp[i]);
		}
		for(int i = 1; i < n - 1; i++) {
			if(ar[i - 1] < ar[i + 1]) {
				maxVal = Math.max(dp[i - 1] + bp[i + 1], maxVal);
			}
		}
		System.out.println(maxVal);
	}
}