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.io.*;
import java.util.*;

public class Main {

	public static void main(String[] args) throws Exception {
		PrintWriter out = new PrintWriter(System.out, true);
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(in.readLine());
		// Scanner s = new Scanner(System.in);
		int n = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(in.readLine());
		int[] mas = new int[n];
		for(int i = 0;i < n;i++) mas[i] = Integer.parseInt(st.nextToken());
		
		int a = 0;
		int b = n-1;
		
		for(int i = 0;i < n-1;i++) {
			if(mas[i] - mas[i+1] == 1) {
				a = i;
				break;
			}
		}
		
		for(int i = a;i < n-1;i++) {
			if(mas[i] - mas[i+1] != 1) {
				b = i;
				break;
			}
		}
		if(a >= b) {
			out.println("0 0");
			return;
		}
		reverse(a,b,mas);
		//out.println(a+" "+b);
		//out.println(Arrays.toString(mas));
		for(int i = 1;i < n;i++) {
			if(mas[i] - mas[i-1] != 1) {
				out.println("0 0");
				return;
			}
		}
		out.println((a+1)+" "+(b+1));
	}
	
	public static void reverse(int l, int r, int[] mas) {
		for(int i = l;i <= (l+r)/2;i++) {
			int c = mas[i];
			mas[i] = mas[r-(i-l)];
			mas[r-(i-l)] = c;
		}
	}

}