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.util.*;
import java.io.*;
import static java.util.Arrays.*;
import static java.lang.Math.*;
import java.math.*;

public class Main {    
    
    void run() throws IOException {
        int n = nint();
        int[] v = new int[n];
        for (int i = 0; i < n; i++) {
            v[i] = nint();
        }
        int a, b;
        for (a = 0; a < n && v[a] == a + 1; a++);
        if (a == n) {
            out.println("0 0");
            return;
        }
        for (b = n - 1; b >= 0 && v[b] == b + 1; b--);
        for (int i = a; i <= b; i++) {
            if (v[i] != b + 1 - (i - a)) {
                out.println("0 0");
                return;
            }
        }
        out.println((a + 1) + " " + (b + 1));
    }
    
    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        //final String FILENAME = "hackercup";
        //in = new BufferedReader(new FileReader(new File(FILENAME + ".in"))); out = new PrintWriter(new File(FILENAME + ".out"));
        in = new BufferedReader(new InputStreamReader(System.in));/* in = new Scanner(System.in);*/ out = new PrintWriter(System.out);
        st = new StringTokenizer(" ");
        new Main().run();
        out.close();
    }

    static BufferedReader in;
    //static Scanner in;
    static PrintWriter out;
    static StringTokenizer st;

    String token() throws IOException {
        while (!st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    int nint() throws IOException {
        return Integer.parseInt(token());
    }

    long nlong() throws IOException {
        return Long.parseLong(token());
    }

    double ndouble() throws IOException {
        return Double.parseDouble(token());
    }
}