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.io.*;
public class Spoilt_Permutation {

	public static void main(String[] args)  throws java.lang.Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(br.readLine()),a[]=new int[n];
		StringTokenizer s = new StringTokenizer(br.readLine());
		for(int i=0; i<n; i++) a[i]=Integer.parseInt(s.nextToken());
		try {
			solve(a);
		}catch(Exception e) {
			System.out.println("# BINOD ");
		}
	}
	static void solve(int a[]) {
		int n=a.length;
		int l=0,r=0;
		for(int i=0; i<n; i++) if(a[i]!=(i+1)) {l=i+1;break;}
		for(int i=n-1; i>=0; --i) if(a[i]!=(i+1)) {r=i+1;break;}
	    if(l==0 || r==0) System.out.println(0+" "+0);
	    else {
	    	boolean ok=true;
			for(int i=l-1,j=r; i<r; i++,j--) {
				if(a[i]!=j) {ok=false;break;}
			}
			System.out.println(l>r || !ok?(0+" "+0):(l+" "+r));
	    }
	}
}