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.PrintWriter; import java.util.Scanner;
  
  
  public class B {
  
  public static void main(String[] args) {
  
  Scanner in = new Scanner(System.in); PrintWriter out = new
  PrintWriter(System.out);
  
  int n = in.nextInt();
  
  int[] a = new int[n+1];
  
  boolean findStart = false, findEnd = false;
  int start = 0;
  int end = 0;
  for (int i=1; i<=n; i++) {
      a[i]=in.nextInt();
      
      if (findStart && (a[i]!=a[i-1]-1) && !findEnd) {
          findEnd = true;
          end = i-1;
      }
      
      if (a[i]!=i && !findStart) {
          findStart = true;
          start = i;
      }
      
  }
  
  if (findStart && !findEnd) {
      findEnd = true;
      end = n;
  }
  
  if (findStart && findEnd) {
      int tmp;
      int start1 = start, end1 = end;
      while (start1<end1) {
          tmp = a[end1];
          a[end1]=a[start1];
          a[start1]=tmp;
          start1++;
          end1--;
      }
      boolean correct = true;
      //System.err.println("im here");
      for (int i=1; i<=n; i++) {
          if (a[i]!=i) {
              correct = false;
              break;
          }
      }
      
      if (correct) {
          out.println(start+" "+end);
      } else {
          out.println("0 0");
      }
  } else {
      out.println("0 0");
  }
  
  in.close(); out.close();
  
  }
  
  }