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 Prob056B
{
    public static void main( String[] Args )
    {
        Scanner scan = new Scanner( System.in );
        int x = scan.nextInt();

        int[] order = new int[x + 1];
        for ( int i = 1; i <= x; i++ )
            order[i] = scan.nextInt();

        int start = 0;
        while ( order[start] == start )
        {
            start++;
            if ( start == x )
                break;
        }

        boolean possible = false;
        if ( start != x )
        {
            possible = true;
            for ( int i = start; i <= order[start]; i++ )
            {
                if ( order[i] + i != order[start] + start )
                {
                    possible = false;
                    break;
                }
            }
            for ( int i = order[start] + 1; i <= x; i++ )
            {
                if ( order[i] != i )
                {
                    possible = false;
                    break;
                }
            }
            
        }

        if ( !possible )
            System.out.println( "0 0" );
        else
            System.out.println( start + " " + order[start] );
    }
}