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.*;

public class Main {

    public static void main(String args[]){
            InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            InputReader in = new InputReader(inputStream);
            PrintWriter out = new PrintWriter(outputStream);
            TaskA solver = new TaskA();
            solver.solve(in,out);
            out.close();
    }

    static class TaskA{
        static int N = 1000;

        public void solve(InputReader in,PrintWriter out){
            int n;
            int a[] = new int[1000+10],b[] = new int[1000+10];
            for (int i = 1;i <= N;i++) b[i] = i;
            n = in.nextInt();
            int l = 0,r = 0;
            for (int i = 1;i <= n;i++)  a[i] = in.nextInt();
            for (int i = 1;i <= n;i++){
                if (a[i]!=b[i]){
                    l = i;
                    break;
                }
            }
            for (int i = n;i >= 1;i--){
                if (a[i]!=b[i]){
                    r = i;
                    break;
                }
            }

            int st = l,en = r;
            while (st<=en){
                int temp = a[st];a[st] = a[en];a[en] = temp;
                st++;en--;
            }
            for (int i = 1;i <= n;i++){
                if (b[i]!=a[i]){
                    out.println("0 0");
                    return;
                }
            }
            out.println(l+" "+r);
        }
    }


    static class InputReader{
        public StringTokenizer st;
        public BufferedReader br;

        public InputReader(InputStream stream){
            br = new BufferedReader(new InputStreamReader(stream));
            st = null;
        }

        public String next(){
            while (st==null || !st.hasMoreTokens()){
                try {
                    st = new StringTokenizer(br.readLine());
                }catch (Exception e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt(){
            return  Integer.parseInt(next());
        }

        public  long nextLong(){
            return  Long.parseLong(next());
        }
    }
}