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.Scanner;

/**
 * @author: zhaoqing
 * @date: 2018/9/9
 * @time: 下午9:37
 */
public class BSpoiltPermutation {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        Integer n = sc.nextInt();
        int start = 10000;
        int end = 10000;
        int t = 0;
        for (int i = 1; i <= n; i++) {
            Integer tmp = sc.nextInt();
            if (start == 10000 && i == tmp) {
                continue;
            } else {
                    if (start == 10000) {
                        start = i;
                        t = tmp;
                        end = i;
                    } else {
                        if ((t - 1) == tmp) {
                            t = tmp;
                            end = i;
                        } else {
                            if (end != 10000) {
                                if (i == tmp) {
                                    continue;
                                } else {
                                    start = 0;
                                    end = 0;
                                    break;
                                }
                            }
                        }
                    }
                }
        }
        if (start == 10000) {
            start = 0;
        }
        if (end == 10000) {
            end = 0;
        }
        System.out.println(start);
        System.out.println(end);
    }
}
