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 Problem{
    public static void main(String args[]){
    Scanner in=new Scanner(System.in);
      int n=in.nextInt();

      int a[]=new int[n+1];

      for(int i=1;i<n+1;i++){
        a[i]=in.nextInt();
      }

      int x = 0;
      for(int i=1;i<n+1;i++){

          if(a[i]!=i)
          {
           x=i;
           break;
          }

      }

      if(x!=0){
        int j=x;
        boolean p=true;

        for(int i=a[x];i>=x;i--){

            if(a[j]!=i){ p=false; break;}
        j++;

        }

        if(p==false) {System.out.println("0 0"); System.exit(0);}



      }
      if(x!=0){
        for(int i=a[x]+1;i<=n;i++){
            if(i!=a[i]) {System.out.println("0 0"); System.exit(0);}

        }

      }
      System.out.println(x+" "+a[x]);

    }

}