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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class B {
    public static void main(String[] args) throws NumberFormatException,
            IOException {

        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        int[] A = new int[n];
        String[] S = in.readLine().split(" ");
        for (int i = 0; i < n; i++)
            A[i] = Integer.parseInt(S[i]);
        int l = 1;
        for (; l < n; l++) {
            if (A[0] != 1) {
                l = 0;
                break;
            }
            if (A[l] - A[l - 1] != 1)
                break;
        }
        int r = n - 2;
        for (; r >= 0; r--) {
            if (A[n - 1] != n) {
                r = n - 1;
                break;
            }
            if (A[r + 1] - A[r] != 1)
                break;
        }
        if (r < l)
            System.out.println("0 0");
        else {
            for (int i = l; i < r; i++) {
                if (A[i] - A[i + 1] != 1) {
                    System.out.println("0 0");
                    return;
                }
            }
            System.out.println((l + 1) + " " + (r + 1));
        }
    }
}
