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 Main {
	static final int MOD_PRIME = 1000000007;

	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Task solver = new Task();

		int i = 0;
		int t = 1;
		// t = in.nextInt();
		for (; i < t; i++)
			solver.solve(i, in, out);

		out.close();
	}

	static class Task {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			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];
			int max = 0;
			dp[n - 1] = 1;
			for (int i = n - 2; i >= 0; i--) {
				if (ar[i + 1] > ar[i]) {
					dp[i] = dp[i + 1] + 1;
				} else {
					dp[i] = 1;
				}
				max = Math.max(max, dp[i]);
			}
			//out.println(max);

			int ptr = 0;
			int m;
			while (ptr < n) {
				m = dp[ptr];
				for (int i = 0; i < m; i++) {
					if (ptr + i + 2 < n && ar[ptr + i + 2] > ar[ptr + i]) {
						max = Math.max(max, m + 1 - dp[ptr + i] + dp[ptr + i + 2]);
						//out.println(ptr + " " + i + " " + max);
					}
				}
				ptr += m;
			}
			out.print(max);

		}

	}

	// template code

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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

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

	}

	static long modexp(long a, long b, long p) {
		// returns a to the power b mod p by modular exponentiation

		long res = 1;
		long mult = a % p;
		while (b > 0) {
			if (b % 2 == 1) {
				res = (res * mult) % p;
			}
			b /= 2;
			mult = (mult * mult) % p;
		}

		return res;
	}

	static double log(double arg, double base) {
		// returns log of a base b, contains floating point errors, dont use for exact
		// calculations.

		if (base < 0 || base == 1) {
			throw new ArithmeticException("base cant be 1 or negative");
		}
		if (arg < 0) {
			throw new ArithmeticException("log of negative number undefined");
		}
		return Math.log10(arg) / Math.log10(base);
	}

	static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}

		return gcd(b, a % b);

	}

	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}

		return gcd(b, a % b);

	}

	// scope for improvement
	static ArrayList<Integer> sieveOfEratosthenes(int n) {
		boolean[] check = new boolean[n + 1];
		ArrayList<Integer> prime = new ArrayList<Integer>();
		for (long i = 2; i <= n; i++) {
			if (!check[(int) i]) {
				prime.add((int) i);
				for (long j = i * i; j <= n; j += i) {
					check[(int) j] = true;
				}
			}
		}

		return prime;
	}

	static int modInverse(int a, int n) {
		// returns inverse of a mod n by extended euclidean algorithm
		int t = 0;
		int newt = 1;
		int r = n;
		int newr = a;
		int quotient;
		int tempr, tempt;
		while (newr != 0) {
			quotient = r / newr;
			tempt = newt;
			tempr = newr;
			newr = r - quotient * tempr;
			newt = t - quotient * tempt;
			t = tempt;
			r = tempr;
		}
		if (r > 1) {
			throw new ArithmeticException("inverse of " + a + " mod " + n + " does not exist");
		} else {
			if (t < 0) {
				t += n;
			}
			return t;
		}
	}

	static int primeModInverse(int a, int n) {
		// returns inverse of a mod n by mod exponentiation, use only if n is prime
		return (int) modexp(a, n - 2, n);
	}
}
