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.*;
import java.util.*;
public class j
{
public static void main(String aa[])throws IOException
{
BufferedReader b=new BufferedReader(new InputStreamReader(System.in));
int k=0,p=0,m=0,i=0;
m=Integer.parseInt(b.readLine());
String s;
int d[]=new int[m];
s=b.readLine();
StringTokenizer c=new StringTokenizer(s);
for(i=0;i<m;i++)
d[i]=Integer.parseInt(c.nextToken());
for(i=0;i<m;i++)
if(d[i]!=i+1)
{
k=i;
break;
}
for(i=m-1;i>=0;i--)
if(d[i]!=i+1)
{
p=i;
break;
}
if(k==p)
{
System.out.print("0 0");
System.exit(0);
}
for(i=k;i<p;i++)
if(d[i+1]>d[i])
{
System.out.print("0 0");
System.exit(0);
}
System.out.print(k+1+" "+(p+1));
}
}