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 RemoveOneElement{

	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		// int test = Integer.parseInt(br.readLine());
		
		// while(test-- >0){

			int n = Integer.parseInt(br.readLine().trim());

			
			
			String s[] = br.readLine().trim().split(" ");

			int a[] = new int[n];
			int c[] = new int[n];
			// HashMap<Integer, Integer> h = new HashMap<Integer, Integer>();


			for(int i=0;i<n;i++){
			    a[i] = Integer.parseInt(s[i]);
			}
			
			// for(int i=0;i<n;i++){
			// 	int start=i;
			// 	while((i+1<n) && a[i]<a[i+1]){
			// 		i++;
			// 	}

			// 	for(int j=start;j<=i;j++){
			// 		c[j] = (i-start)+1;
			// 	}
			// 	max = Math.max(max, (i-start+1));
			// }

			
			// for(int i=0;i<n;i++){
			// 	// System.out.print(c[i] + " ");
			// 	if(i+1<n && c[i]!=c[i+1]){
			// 		if(i-1<=0 || a[i-1]<a[i+1]){
			// 			// System.out.println("a " + a[i-1]+ " " + a[i+1]);
			// 			max = Math.max(max, c[i]+c[i+1]-1);
			// 		}else if(i+2>=n || a[i]<a[i+2]){
			// 			// System.out.println("b " + a[i]+ " " + a[i+2]);
			// 			max = Math.max(max, c[i]+c[i+1]-1);
			// 		}
			// 	}
			// }

			int start[] = new int[n];
			int end[] = new int[n];
			int max = 0;
			Arrays.fill(start,1);
			Arrays.fill(end,1);
			for(int i=1;i<n;i++){
				if(a[i]>a[i-1])
					end[i]=end[i-1]+1;
				max = Math.max(max, end[i]);
			}
			for(int i=n-2;i>=0;i--){
				if(a[i]<a[i+1])
					start[i] = start[i+1]+1;
				max = Math.max(max, start[i]);
			}

			
			for(int i=1;i<n-1;i++){
				if(a[i-1]<a[i+1]){
					max = Math.max(max, end[i-1] + start[i+1]);
				}
			}

			System.out.println(max);

		// }
	}
}