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.BufferedInputStream;
import java.text.DateFormatSymbols;
import java.util.Scanner;


public class Main {



static int[] a=new int[10000];



	public static void main(String[] args) {
		Scanner scanner = new Scanner(new BufferedInputStream(System.in));
      
		int n=scanner.nextInt();
		for(int i=1;i<=n;i++)a[i]=scanner.nextInt();
		int i=1;
		int p=0;
		 while(a[i]==i&&i<=n)i++;
		 int j=n;
		 while(a[j]==j&&j>1)j--;
		 if(i<j) {
			 for(int k=i;k<j;k++)if(a[k]!=a[k+1]+1)p=1;
		 }
		 else p=1;
		 System.out.println(p==1?"0 0":(i+" "+j));
	}
}