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

public class Z implements Runnable{

    BufferedReader in;
    StringTokenizer st;
    PrintWriter out;
    
    void solve() throws IOException{
        int n = nextInt(), pos1 = -1, pos2 = -1;
        int[] a = new int[n], b = new int[n];
        
        for(int i = 0; i < n; i++) {
            a[i] = nextInt();
            b[i] = i + 1;
        }
        
        for(int i = 0; i < n; i++) {
            if(a[i] != b[i]) {
                pos1 = i;
                while(i < n - 1 && a[i] == a[i + 1] + 1) {
                    i++;
                    pos2 = i + 1;
                }
                break;
            }
        }
        if(pos1 == - 1 || pos2 == -1) {
            out.print("0 0");
            return;
        }
        for(int i = pos1; i < pos2; i++) {
            if(pos2 - i + pos1 - 1 > i) {
                int tmp = a[pos2 - i + pos1 - 1];
                a[pos2 - i + pos1 - 1] = a[i];
                a[i] = tmp;
            }
        }
        for(int i = 0; i < n; i++)
            if(a[i] != b[i]) {
                out.print("0 0");
                return;
            }
        out.print((pos1 + 1) + " " + pos2);
    }
    
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            eat("");
            solve();
            out.close();
            in.close();
        } catch (IOException ex) {
            throw new RuntimeException(ex);
        }
    }
    
    void eat(String s) {
        st = new StringTokenizer(s);
    }
    
    String next() throws IOException {
        while (!st.hasMoreTokens()) {
            String line = in.readLine();
            if (line == null) {
                return null;
            }
            eat(line);
        }
        return st.nextToken();
    }
    
    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    
    long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    
    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
    
    public static void main(String[] args) {
        Locale.setDefault(Locale.US);
        new Thread(new Z()).start();
    }

}