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.*;
import java.lang.*;
import java.io.*;




/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
     
    static PrintWriter out;
    
    
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			out=new PrintWriter(System.out);
		}
		String next(){
			while(st==null || !st.hasMoreElements()){
				try{
					st= new StringTokenizer(br.readLine());
				}
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		long nextLong(){
			return Long.parseLong(next());
		}
		double nextDouble(){
			return Double.parseDouble(next());
		}
		String nextLine(){
			String str = "";
			try{
				str=br.readLine();
			}
			catch(IOException e){
				e.printStackTrace();
			}
			return str;
		}
	}
	

// SHIVAM GUPTA :



 
 
    
    
public static void main (String[] args) throws java.lang.Exception
{
   FastReader scn = new FastReader() ;

int n = scn.nextInt() ;
int[] arr = new int[n +1];


for(int i = 1; i < n + 1 ; i++)arr[i] = scn.nextInt() ;

int index = -1  ;

for(int i = 1; i < n+1  ; i++ )
{
       if(arr[i] != i)
       {
         index = i ; 
         break ;
       }
       
       
}

if(index ==-1)out.println(0 + " " + 0) ;

else{

int j = arr[index]  ;

int in = index;
int val = j ;


while(in < val)
{
   int temp = arr[in] ;
   arr[in] = arr[val] ;
   arr[val] = temp ;
       
   val-- ;
  in++ ;
       
}
boolean ans = true  ;

for(int i =1; i < n+1  ;i++)
{
       if(arr[i] != i){
              ans = false  ;break ;
       } 
}


if(ans ==true)
{
       out.println(index + " " + j ) ;
}

else{
       
       out.println(0 + " " + 0) ;
       
}



}
  
  
out.flush()  ; 
	}
	
	
}