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.*;
public class AB {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int r = sc.nextInt();
		boolean reverse = true;
		int k=0;
		int start=0;
		int end=-1;
		int counter=0;
		int j=0;
		for(int i=1;i<r+1;i++)
		{
			
			if(k!=0)
				j=k;
			else
				j=i;
			int a = sc.nextInt();
			if(a!=j)
			{
				//System.out.println(a+" "+counter);
				if(reverse==false){
					break;
				}
				if(k==0){
					k=a+1;
					end=k;
					start=i;
				}
				
				if(a==k-1-counter)
					counter++;
				else
					reverse=false;
			}
			else
				k=0;
			
		}
		if(reverse && end>0)
		System.out.println(start+" " + (end-1));
		else
			System.out.println(0+" "+0);
		
	}
}