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.*;
import java.util.*;
import java.math.*;

public class Solution {

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer st;

    public String next() throws Exception {
        if (st == null || !st.hasMoreElements()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws Exception {
        return Integer.parseInt(next());
    }
    public long nextLong() throws Exception {
        return Long.parseLong(next());
    }
    public double nextDouble() throws Exception {
        return Double.parseDouble(next());
    }

    public void run() throws Exception {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        int n = nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; ++i) {
            a[i] = nextInt();
        }
        int s = 0, e = 0;
        for (int i = 0; i < n - 1; i++) {
            if (i != a[i] - 1) {
                if (a[i] > a[i + 1]) {
                    s = i + 1;
                    while(i + 1 < n && a[i] - a[i + 1] == 1) ++i;
                    e = i + 1;
                    ++i;
                    while(i < n) {
                        if (a[i] != i + 1) {
                            out.println(0 + " " + 0);
                            out.close();
                            return;
                        }
                        ++i;
                    }

                } else {
                    out.println(0 + " " + 0);
                    out.close();
                    return;
                }
            }
        }
        out.println(s + " " + e);
        out.close();
    }

    public static void main(String[] args) throws Exception {
        new Solution().run();
    }
}
