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.math.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class P056B{
	Scanner sc=new Scanner(System.in);

	int INF=1<<28;
	double EPS=1e-9;

	int n;
	int[] a;

	void run(){
		n=sc.nextInt();
		a=new int[n];
		for(int i=0; i<n; i++){
			a[i]=sc.nextInt();
		}
		solve();
	}

	void solve(){
		for(int i=0; i<n; i++){
			a[i]--;
		}
		int left;
		for(left=0; left<n&&a[left]==left; left++);
		int right;
		for(right=n-1; right>=0&&a[right]==right; right--);
		// debug(left, right);
		boolean ok=left<right;
		for(int i=left; i<=right; i++){
			ok&=a[i]==-i+left+right;
		}
		// debug(ok);
		if(ok){
			println((left+1)+" "+(right+1));
		}else{
			println("0 0");
		}
	}

	void println(String s){
		System.out.println(s);
	}

	void print(String s){
		System.out.print(s);
	}

	void debug(Object... os){
		System.err.println(deepToString(os));
	}

	public static void main(String[] args){
		Locale.setDefault(Locale.US);
		new P056B().run();
	}
}