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
/**
 * Created by IntelliJ IDEA.
 * User: aircube
 * Date: 11.01.11
 * Time: 4:14
 * To change this template use File | Settings | File Templates.
 */

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


public class TaskB {
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;

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

    void run() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        st = null;
        out = new PrintWriter(System.out) ;

        solve();

        br.close();
        out.close();
    }
    void solve() throws  IOException {
        int n =  nextInt();
        int a[] = new int[n];
        for(int i = 0; i < n; ++i)
            a[i] = nextInt();
        int b[] = new int[n];
        for(int i = 0; i < n; ++i) {
            b[i] = a[i] - (i + 1);
        }
        int cntZero = 0;
        for(int i = 0; i < n; ++i) {
            if (b[i] != 0) ++ cntZero;
        }
        if (cntZero < 2) {
            out.print("0 0");
            return ;
        }
        for(int i = 0; i < n; ++i) {
            if (b[i] != 0) {
                int p = i + 1;
                while (p < n && a[p - 1] - a[p] == 1) ++ p;
                for(int u = p; u < n; ++u)
                    if (b[u] != 0) {
                        out.print("0 0");
                        return ;
                    }
                out.print((i + 1) + " " + p);
                return;
            }
        }

    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
    BigInteger nextBigInteger() throws IOException {
        return new BigInteger(nextToken());
    }
    String nextToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }
}