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

public class Main {
	
	static int arr[] = new int[1010], n;
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		n = scan.nextInt();
		
		int l = Integer.MAX_VALUE;
		int r = Integer.MIN_VALUE;
		
		for(int i = 1; i <= n; i ++) {
			arr[i] = scan.nextInt();
			if(arr[i] != i) {
				l = min(l, i);
				r = max(r, i);
			}
		}
		
		if(check(l,r)) {
			System.out.println(l + " " + r);
		} else {
			System.out.println("0 0");
		}
		
	}

	private static boolean check(int l, int r) {
		if(l == Integer.MAX_VALUE || r == Integer.MIN_VALUE)	return false;
		for(int i = 1; i <= n; i ++) {
			if(i < l || i > r) {
				if(arr[i] != i)	return false;
			}
		}
		for(int i = l; i <= r; i ++) {
			if(arr[i] != l+r-i)	return false;
		}
		return true;
	}
}