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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class SpoiltPermutation {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        ArrayList<Integer> arrayList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            arrayList.add(s.nextInt());
        }
        System.out.println(isSortable(arrayList));
    }

    private static String isSortable(ArrayList<Integer> arrayList) {
        ArrayList<Integer> sortedArray = new ArrayList<>(arrayList);
        Collections.sort(sortedArray);
        int start = -1, end = -1;
        for (int i = 0; i < arrayList.size(); i++) {
            if (!arrayList.get(i).equals(sortedArray.get(i))) {
                if (start == -1) {
                    start = i;
                } else {
                    end = i;
                }
            }
        }

        ArrayList<Integer> subArray = new ArrayList<>();
        if (start == -1) {
            return "0 0";

        }
        for (int i = start; i <= end; i++) {
            subArray.add(arrayList.get(i));
        }
        Collections.reverse(subArray);
        for (int i = 0; i < start; i++) {
            if (!arrayList.get(i).equals(sortedArray.get(i))) {
                return "0 0";
            }
        }
        for (int i = 0; i < subArray.size(); i++) {
            if (!sortedArray.get(i + start).equals(subArray.get(i))) {
                return "0 0";
            }
        }
        for (int i = end + 1; i < arrayList.size(); i++) {
            if (!arrayList.get(i).equals(sortedArray.get(i))) {
                return "0 0";
            }
        }
        return (start + 1) + " " + (end + 1);


    }
}