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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.math.BigInteger;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader sc = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(1, sc, out);
        out.close();
    }

    static class Task {
        public void solve(int testNumber, InputReader sc, PrintWriter out) throws IOException {
        	int n=sc.nextInt();
        	int[] a=new int[n+2];
        	int ans=1;
        	int[] pre=new int[n+1];
        	int[] suf=new int[n+2];
        	pre[1]=suf[n]=1;
        	
        	for(int i=1;i<=n;i++)
        		a[i]=sc.nextInt();
        	for(int i=2;i<=n;i++) {
        		if(a[i]>a[i-1])
        			pre[i]=pre[i-1]+1;
        		else
        			pre[i]=1;
        	}
        	for(int i=n-1;i>=1;i--) {
        		if(a[i]<a[i+1]) 
        			suf[i]=suf[i+1]+1;
        		else
        			suf[i]=1;
        	}
        	
        	for(int i=1;i<=n;i++) {
        		ans=Math.max(ans, Math.max(pre[i], suf[i]));
        		if(a[i-1]<a[i+1])
        			ans=Math.max(ans, pre[i-1]+suf[i+1]);
        	}
        	out.println(ans);
        }

    }

    static class InputReader{
        StreamTokenizer tokenizer;
        public InputReader(InputStream stream){
            tokenizer=new StreamTokenizer(new BufferedReader(new InputStreamReader(stream)));
            tokenizer.ordinaryChars(33,126);
            tokenizer.wordChars(33,126);
        }
        public String next() throws IOException {
            tokenizer.nextToken();
            return tokenizer.sval;
        }
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }
        public boolean hasNext() throws IOException {
            int res=tokenizer.nextToken();
            tokenizer.pushBack();
            return res!=tokenizer.TT_EOF;
        }
        
        public double nextDouble() throws NumberFormatException, IOException {
        	return Double.parseDouble(next());
        }
        
        public BigInteger nextBigInteger() throws IOException {
        	return new BigInteger(next());
        }
    }
}
