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.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.Scanner;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Reader;
import java.io.Writer;
import java.io.InputStream;

public class B56 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), l = 0, r = n - 1;
		int[] v = new int[n];
		boolean flag = false;
		for (int i = 0; i < n; i++) {
			v[i] = in.nextInt() - 1;
			if (!flag && i == v[i]) {
				l++;
			} else {
				flag = true;
			}
		}
		flag = false;
		for (int i = n - 1; i >= 0 && !flag; i--) {
			if (i == v[i]) {
				r--;
			} else {
				flag = true;
			}
		}
		if (l == n && r == -1) {
			System.out.println("0 0");
			return;
		}
		int a = l, b = r;
		while (l < r) {
			if (v[l] != r || v[r] != l) {
				System.out.println("0 0");
				return;
			}
			l++; r--;
		}
		System.out.println((a + 1) + " " + (b + 1));
	}
}

