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

public class Main implements Runnable {
    static StringTokenizer st;
    static BufferedReader br;
    static PrintWriter out;
    static Random rand;

    public static void main(String[] args) throws IOException {
	rand = new Random();
	br = new BufferedReader(new InputStreamReader(System.in));
	out = new PrintWriter(new OutputStreamWriter(System.out));
	new Main().run();
	out.flush();
    }

    String nline() {
	try {
	    return br.readLine();
	} catch (Exception e) {
	    return null;
	}
    }

    String ns() {
	try {
	    while (st == null || !st.hasMoreTokens())
		st = new StringTokenizer(br.readLine());
	    return st.nextToken();
	} catch (Exception e) {
	    return null;
	}
    }

    int ni() {
	return Integer.valueOf(ns());
    }

    long nl() {
	return Long.valueOf(ns());
    }

    double nd() {
	return Double.valueOf(ns());
    }

    public void run() {
	Solve();
    }

    void Solve() {
	int n = ni();
	int[] mas = new int[n];
	int f = -1, t = -1;
	for (int i = 0; i < n; i++) {
	    mas[i] = ni();
	    if (mas[i] != i + 1 && f == -1)
		f = i;
	    if (mas[i] != i + 1)
		t = i;
	}
	boolean bad = false;
	for (int i = f; i < t; i++) {
	    if (mas[i] != mas[i + 1] + 1) {
		bad = true;
		break;
	    }
	}
	if (bad) {
	    out.println("0 0");
	} else {
	    if (f == -1 && t == -1) {
		out.println("0 0");
	    } else {
		out.println((f + 1) + " " + (t + 1));
	    }
	}
    }
}
