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 B56 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0 ; i < n ; i++) {
            a[i] = in.nextInt() - 1;
        }
        for (int i = 0 ; i < n ; i++)
            if (i != a[i]) {
                int l = i;
                int r = a[i];
                for (int j = l ; j <= (l + r) / 2 ; j++) {
                    int tmp = a[j];
                    a[j] = a[r - j + l];
                    a[r - j + l] = tmp;
                }
                boolean ok = true;
                for (int j = 1 ; j < n ; j++)
                    ok &= (a[j] == a[j - 1] + 1);
                if (ok) {
                    l++;
                    r++;
                    System.out.println(l + " " + r);
                    System.exit(0);
                } else {
                    System.out.println("0 0");
                    System.exit(0);
                }
            }
        System.out.println("0 0");
    }
}