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

public class Template {
	
	static Scanner sc = new Scanner(System.in);
	static InputStreamReader stream = new InputStreamReader(System.in);
	static BufferedReader br = new BufferedReader(stream);
	

	public static void main(String[] args) throws IOException{
		
		int n = readInt();
		int[] a = readArrayline(n);
		int xi=-1, yi=-1;
		
		for(int i=0 ; i<n ; i++) {
			if(a[i]!=i+1) {
				xi=i;
				break;
			}
		}
		
		for(int i=n-1 ; i>=0 ; i--) {
			if(a[i]!=i+1) {
				yi=i;
				break;
			}
		}
		
		if(function(a,xi,yi)) {
			System.out.println(xi+1 + " " + a[xi]);
		}else {
			System.out.println("0 0");
		}
	}
	
	public static boolean function(int[] a , int xi, int yi) {
		if(xi==-1) return false;
		int x = a[xi];
		for(int i=xi ; i<=yi ; i++) {
			if(a[i] != x-(i-xi)) return false;
		}
		return true;
	}
	
	//BufferedReader Input functions
	public static int readInt() throws IOException{
		return (Integer.parseInt(br.readLine()));
	}
	
	public static int[] readArray(int n) throws IOException{
		int[] a = new int[n];
		for(int i=0 ; i<n ; i++) {
			a[i]= readInt();
		}
		return a;
	}
	
	public static int[] readArrayline(int n) throws IOException{
		int[] a = new int[n];
		String[] s = br.readLine().split(" ");
		for(int i=0 ; i<n ; i++) {
			a[i]= Integer.parseInt(s[i]);
		}
		return a;
	}

}