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 Main {
	public static void main(String[] args) throws Exception {
		MScanner sc = new MScanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n=sc.nextInt();
		int[]in=sc.takearr(n);
		int[][]seq=new int[n][2];
		seq[0][0]=1;
		for(int i=1;i<n;i++) {
			seq[i][0]=1;
			
			if(in[i]>in[i-1]) {
				seq[i][0]=Math.max(seq[i][0], seq[i-1][0]+1);
				seq[i][1]=Math.max(seq[i][1], seq[i-1][1]+1);
			}
			if(i>=2 && in[i]>in[i-2]) {
				seq[i][1]=Math.max(seq[i][1], seq[i-2][0]+1);
			}
		}
		int max=0;
		for(int i[]:seq){
			max=Math.max(max, Math.max(i[0], i[1]));
			//System.out.println(i[0]+" "+i[1]);
		}
		pw.println(max);
		pw.flush();
	}
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;

		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int[] takearr(int n) throws IOException {
			int[] in = new int[n];
			for (int i = 0; i < n; i++)
				in[i] = nextInt();
			return in;
		}

		public long[] takearrl(int n) throws IOException {
			long[] in = new long[n];
			for (int i = 0; i < n; i++)
				in[i] = nextLong();
			return in;
		}

		public Integer[] takearrobj(int n) throws IOException {
			Integer[] in = new Integer[n];
			for (int i = 0; i < n; i++)
				in[i] = nextInt();
			return in;
		}

		public Long[] takearrlobj(int n) throws IOException {
			Long[] in = new Long[n];
			for (int i = 0; i < n; i++)
				in[i] = nextLong();
			return in;
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}