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

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

class TaskB {
    public void solve(int testNumber, Reader in, PrintWriter out) {
        int n = in.ni();
        int[] a = new int[n];
        Helper.fillIntArray(in, a);
        for(int i = 0; i < n; i++) {
            if(a[i] == i + 1) {
                continue;
            } else {
                for(int j = i; j < a[i] - 1; j++) {
                    if(a[j + 1] - a[j] != -1) {
                        out.println(0 + " " + 0);
                        return;
                    }
                }
                for(int j = a[i]; j < n; j++) {
                    if(a[j] != j + 1) {
                        out.println(0 + " " + 0);
                        return;
                    }
                }
                out.println(i + 1 + " " + a[i]);
                return;
            }
        }

        out.println(0 + " " + 0);
    }
}

class Reader {
    private BufferedReader in;
    private StringTokenizer st = new StringTokenizer("");
    private String delim = " ";

    public Reader(InputStream in) {
        this.in = new BufferedReader(new InputStreamReader(in));
    }

    public String next() {
        if (!st.hasMoreTokens()) {
            st = new StringTokenizer(rl());
        }
        return st.nextToken(delim);
    }

    public String rl() {
        try {
            return in.readLine();
        } catch(IOException e) {
            throw new RuntimeException(e);
        }
    }

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

}

class Helper {

    public static void fillIntArray(Reader in, int[] a) {
        for(int i = 0; i < a.length; i++) {
            a[i] = in.ni();
        }
    }

}

