Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.

Input

The first line contains an integer n (1 ≤ n ≤ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.

Output

If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 ≤ l < r ≤ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.

Examples

Input

8
1 6 5 4 3 2 7 8


Output

2 6


Input

4
2 3 4 1


Output

0 0


Input

4
1 2 3 4


Output

0 0
import java.io.*;
import java.util.*;

public class Main implements Runnable {

	public void _main() throws IOException {
		int n = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt() - 1;
		for (int l = 0; l < n; l++)
			for (int r = l + 1; r < n; r++)
				if (can(a, n, l, r)) {
					out.print((l + 1) + " " + (r + 1));
					return;
				}
		out.print("0 0");
	}

	private boolean can(int[] a, int n, int l, int r) {
		for (int i = 0; i < l; i++)
			if (a[i] != i)
				return false;
		for (int i = r + 1; i < n; i++)
			if (a[i] != i)
				return false;
		for (int i = l, j = 0; i <= r; i++, j++)
			if (a[l + j] != r - j)
				return false;
		return true;
	}

	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer st;

	private String next() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String rl = in.readLine();
			if (rl == null)
				return null;
			st = new StringTokenizer(rl);
		}
		return st.nextToken();
	}

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

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

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

	public static void main(String[] args) {
		Locale.setDefault(Locale.UK);
		new Thread(new Main()).start();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);

			_main();

			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(202);
		}
	}

}
