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.*;
import java.util.*;
public class RemoveElement {

	public static void main(String[] args) throws IOException{
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int arr[] = new int[N];
		for(int i = 0; i < N; ++i) {
			arr[i] = Integer.parseInt(st.nextToken());
		}
		int pSum[] = new int[N], sSum[] = new int[N];
		pSum[0] = 1; sSum[N-1] = 1; 
		for(int i = 1; i < N; ++i) {
			if(arr[i]>arr[i-1]) {
				pSum[i] = pSum[i-1]+1;
			}else {
				pSum[i] = 1;
			}
		}
		int ans = 1;
		for(int i = N-2; i >= 0; --i) {
			if(arr[i] < arr[i+1]) {
				sSum[i] = sSum[i+1]+1;
			}else {
				sSum[i] = 1;
			}
			ans = Math.max(ans, sSum[i]);
		}
		for(int i = 0; i < N-2; ++i) {
			if(arr[i] < arr[i+2]) {
				ans = Math.max(pSum[i]+sSum[i+2], ans);
			}
		}
		System.out.println(ans);
	}

}
