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;

/**
 * @author Alexander Grigoryev
 *         Created on 27.07.2011
 */
public
class Main {
    static Scanner in = new Scanner(System.in);

    public static
    void main(String[] args) {
        int n = in.nextInt();

        int[] a = new int[n];
        for(int i = 0; i < n; i++) a[i] = in.nextInt();

        int ct = 0, l = 0, r = 0;
        for(int i = 0; i < n; i++) {
            if(a[i] != i + 1) {
                ct++;
                int j = i + 1;
                for(; j < a[i]; j++) {
                    if(a[j] != a[j - 1] - 1) fail();
                }
                l = i + 1;
                r = j;
                i = j;
            }
        }
        System.out.println(ct == 1 ? l + " " + r : "0 0");
    }

    static
    void fail() {
        System.out.println("0 0");
        System.exit(0);
    }
}