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.util.*;


public class B
{
   public static void main(String[] args)
   {
      new B(new Scanner(System.in));
   }

   public B(Scanner in)
   {
      int N = in.nextInt();
      int[] vs = new int[N];
      int[] rs = new int[N];
      for (int i=0; i<N; i++)
         rs[i] = vs[i] = in.nextInt()-1;

      int ri = -1;
      int rj = -1;
      for (int i=0; i<N; i++)
         if (vs[i] != i)
         {
            for (int j=i; j<=vs[i]; j++)
               rs[j] = vs[vs[i]+i-j];

            ri = i;
            rj = vs[i];
            break;
         }

      boolean passed = true;
      for (int i=0; i<N; i++)
         if (rs[i] != i)
            passed = false;
      
      if (passed)
         System.out.printf("%d %d%n", ri+1, rj+1);
      else
         System.out.println("0 0");
   }
}

