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.util.Scanner;


public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
     Scanner sc=new Scanner(System.in);
     int n=sc.nextInt();
     int[] mass=new int[n];
     for(int i=0; i<n; i++)
         mass[i]=sc.nextInt();
     int l=-1,r=-1;
     boolean ok=true;
     for(int i=0; i<n; i++)
     {
         if(mass[i]==i+1)
             continue;
         if(l==-1)
         {
             if(mass[i]!=i+1)
             {
                 if(l!=-1)
                 {
                     ok=false;
                     break;
                 }
                 l=i;
             }
         }
         else
         { 
             while(i<n && mass[i]==mass[i-1]-1)
                 i++;
             
                 if(r!=-1)
                 {
                     ok=false;
                     break;
                 }
                 r=i;
             }
             }
         
         
     if(l==-1 || r==-1 || !ok)
     {
         System.out.print("0 0");
     }
     else
         System.out.print((l+1)+" "+r);
     }
    }


///ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE