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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {

	void solve() throws IOException {
		int n = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < a.length; i++) {
			a[i] = nextInt();
		}
		for (int i = 0; i < a.length; i++) {
			if (a[i] != i + 1) {
				// out.println("BAD at " + (i+1));
				// out.println("CHECKING " + (i+1) + " to " + a[i]);
				for (int j = i; j < a[i]; j++)
					if (a[j] != a[i] - (j - i)) {
						out.println("0 0");
						return;
					}
				int l = i + 1;
				int r = a[i];
				for (int j = a[i] + 1; j < n; j++)
					if (a[j] != j + 1) {
						out.println("0 0");
						return;
					}
				out.println(l + " " + r);
				return;
			}
		}
		out.println("0 0");
	}

	BufferedReader in;
	StringTokenizer st;
	PrintWriter out;

	public void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		eat("");
		solve();
		out.close();
		in.close();
	}

	void eat(String s) {
		st = new StringTokenizer(s);
	}

	String next() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return null;
			}
			eat(line);
		}
		return st.nextToken();
	}

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

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

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

	public static void main(String[] args) throws IOException {
		new Main().run();
	}
}