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.io.BufferedInputStream;
import java.util.Scanner;

/**
 * Created by jizhe on 2016/1/9.
 */
public class SpoiltPermutation {
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));

        int N = in.nextInt();
        int l = 0, r = 0, count = 0;
        for( int i = 1; i < N+1; i++ )
        {
            int number = in.nextInt();
            if( l == 0 )
            {
                if( number != i )
                {
                    l = i;
                    r = number;
                    count++;
                }
            }
            else
            {
                //System.out.printf("l: %d, r: %d, count: %d, r-l+1: %d, r-count+1: %d\n", l, r, count, r-l+1, r-count+1);
                if( count <= r-l )
                {
                    if( number != r-count )
                    {
                        System.out.printf("0 0\n");
                        return;
                    }
                    count++;
                }
                else
                {
                    if( number != i )
                    {
                        System.out.printf("0 0\n");
                        return;
                    }
                }
            }
        }

        System.out.printf("%d %d\n", l, r);
    }
}
