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.Scanner;
public class Ishu
{
	public static void main(String[] args)
	{
	Scanner scan=new Scanner(System.in);
	int n,i,sum=0,sm=0,big=0;
	int[] a=new int[1000];
	n=scan.nextInt();
	for(i=0;i<n;++i)
		a[i]=scan.nextInt();
	if(a[0]==n&&a[n-1]==1&&n>1)
		{
		for(i=0;i<n-1;++i)
			if(!(a[i]-a[i+1]==1))
				{
				System.out.println("0 0");
				break;
				}
		if(i==n-1)
			System.out.println(1+" "+n);
		}	
	else if(!(a[0]==1)&&a[n-1]==n)
		{
		for(i=0;i<n-1;++i)
			if(a[i]-a[i+1]<-1)
				++sum;
		if(sum>1)
			System.out.println("0 0");
		else 
			System.out.println(1+" "+a[0]);
		}
	else if(!(a[n-1]==n)&&a[0]==1)
		{
		for(i=0;i<n-1;++i)
			if(a[i]-a[i+1]<-1)
				++sum;
		if(sum>1)	
			System.out.println("0 0");
		else
			System.out.println(a[n-1]+" "+n);
		}
	else if(a[0]==1&&a[n-1]==n)
		{
		for(i=0;i<n-1;++i)
			if(a[i]-a[i+1]<-1)
				{
				if(sum==0)
					sm=i+1;
				else if(sum==1)
					big=i;
				++sum;
				}
		if(sum>2||sum==0)
			System.out.println("0 0");
		else 
			System.out.println(a[big]+" "+a[sm]);
		}
	else
		System.out.println("0 0");
	}
}