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.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;


public class Main {
    
    public static void main(String[] args) throws Exception {
        
        Main main = new Main();
        main.doWork();
        
    }
    
    private void doWork() throws Exception
    {       
        StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        
        st.nextToken();
        int n = (int) st.nval;
        
        int left = -1, right = -1;
        
        int array[] = new int[n];
        
        for (int i=0; i<n; i++)
        {
            st.nextToken();
            array[i] = (int) st.nval;
        }
        
        //find left end
        for (int i=0; i<n-1; i++)
        {
            if (array[i+1] < array[i])
            {
                left = i;
                break;
            }
        }
        
        if (left != -1)
        {
            for (int i=left; i<n-1; i++)
            {
                if (array[i+1] > array[i])
                {
                    right = i;
                    break;
                }
            }
            
            if (right == -1)
                right = n-1;
        }
        
        boolean ok = true;
    
        if (left == -1 || right == -1)
        {
            ok = false;
        } else
        {
            //now reverse the middle
            int rright = right;
            for (int i=left; i<rright; i++)
            {
                int temp = array[i];
                array[i] = array[rright];
                array[rright] = temp;
                
                rright--;
            }
            
            //check if ok
            if (array[0] != 1)
                ok = false;
            
            for (int i=1; i<n; i++)
                if (array[i] != array[i-1] + 1)
                    ok = false;
        }
        
        //1 based indexing
        left++; right++;
        
        if (ok == false)
            pw.write("0 0\n");
        else
            pw.write(left + " " + right + "\n");
                    
        pw.flush();
        pw.close();
    }   
}
