Problem: 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

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.StringTokenizer;

/**
 * _56B
 * θ(n) time
 * θ(1) space
 *
 * @author artyom
 */
public class _56B implements Runnable {
    private BufferedReader in;
    private StringTokenizer tok;

    private Object solve() throws IOException {
        int z = -1, t = -1, q = 0;
        for (int i = 1, n = nextInt(); i <= n; i++) {
            int x = nextInt();
            if (q == 0) {
                if (x != i) {
                    z = i;
                    t = x;
                    q = 1;
                }
            } else if (q == 1) {
                if (x != t + z - i) {
                    return "0 0";
                }
                if (i == t) {
                    q = 2;
                }
            } else {
                if (x != i) {
                    return "0 0";
                }
            }
        }
        return q == 2 ? z + " " + t : "0 0";
    }

    //--------------------------------------------------------------
    public static void main(String[] args) {
        new _56B().run();
    }

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            tok = null;
            PrintStream out = System.out;
            out.print(solve());
            in.close();
        } catch (IOException e) {
            System.exit(0);
        }
    }

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

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

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