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
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            int[] a=new int[1010];
            int n=scanner.nextInt();
            int l=0,r=0;
            for(int i=1;i<=n;i++)
                a[i]=scanner.nextInt();
            for(int i=1;i<=n;i++)
                if(a[i]!=i)
                {
                    l=i;
                    break;
                }
            for(int i=n;i>=1;i--)
                if(a[i]!=i)
                {
                    r=i;
                    break;
                }
            for(int i=l;i<r;i++)
            {
                if(a[i]<a[i+1])
                {
                    System.out.println("0 0");
                    return ;
                }
            }
            if(l==0&&r==0)
            {
                System.out.println("0 0");
                return;
            }
            System.out.println(l+" "+r);
        }
    }
}
 	 		 	  			 						 		   					