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 javax.print.attribute.HashDocAttributeSet;
import java.io.*;
import java.util.*;

public class Main{
    public static void main(String[] args) {
        MyScanner scanner = new MyScanner();
        int n = scanner.nextInt(), arr[] = new int[ n ];

        for( int i = 0;  i < n ; i++ )
        {
            arr[ i ] = scanner.nextInt();
        }
        int f = -1, t = -1;
        for( int i = 1 ; i < n ; i++ )
        {
            if( arr[ i - 1 ] > arr[ i ] )
            {
                f = i - 1;
                break;
            }
        }
        for( int i = n - 1 ; i > 0 ; i-- )
        {
            if( arr[ i - 1 ] > arr[ i ] )
            {
                t = i;
                break;
            }
        }
        if( f > t - 1 || f == -1 || t == -1 )
        {
            System.out.println("0 0");
            return;
        }
        int len = ( ( t - f ) )/ 2;
        for( int i = 0 ; i <= len ; i++ )
        {
            int temp = arr[f + i];
            arr[ f + i ] = arr[ t - i];
            arr[ t - i] = temp;

        }
        for( int i = 0 ; i < n - 1; i++ )
        {
            if( arr[i] > arr[i + 1 ])
            {
                System.out.println("0 0");
                return;
            }

        }
        System.out.println((f + 1)+ " " + (t+1));
    }
}

class MyScanner {
    BufferedReader br;
    StringTokenizer st;

    public MyScanner() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public int mod(long x) {
        // TODO Auto-generated method stub
        return (int) x % 1000000007;
    }

    public int mod(int x) {
        return x % 1000000007;
    }

    boolean hasNext() {
        if (st.hasMoreElements())
            return true;
        try {
            st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
        return st.hasMoreTokens();
    }

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

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

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
    public int[] readIntArray( int n )
    {
        int ar[] = new int[ n ];
        for( int i = 0;  i < n ; i++ )
            ar[ i ] = this.nextInt();
        return ar;
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
}