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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;


public class problemB implements Runnable{
	
	public void solve() throws IOException{
		int n=intArr()[0];
		int[] x=intArr();
		
		int l=0,r=0;
		boolean found=false;
		boolean allCorrect=true;
		
		for(int i=1;i<=n;i++){
			int xx=x[i-1];
			if(found && i>=l && i<=r){
				if(xx==r-(i-l)){
					continue;
				} else {
					allCorrect=false;
					break;
				}
			} else {
				if(xx==i){
					continue;
				} else if(!found){
					found=true;
					l=i;
					r=xx;
					continue;
				} else {
					allCorrect=false;
					break;
				}
			}
		}
		
		if(!(found && allCorrect)){
			out.println("0 0");
		} else {
			out.print(l);
			out.print(" ");
			out.println(r);
		}
	}

    public static void main(String[] args){
		new problemB().run();
    }

	@Override
	public void run(){
//		System.out.println("trying to solve");
		try{
			in=new BufferedReader(new InputStreamReader(stdin));
			out=new PrintWriter(stdout);
			solve();
			in.close();
			out.close();
		} catch (Exception ex){
			ex.printStackTrace();
			System.exit(1);
		}
	}
	
	BufferedReader in;
	PrintWriter out;
	
	int[] intArr() throws IOException {
		String[] arr=nextLine().split("\\s");
		int[] res=new int[arr.length];
		for(int i=0;i<arr.length;i++) res[i]=Integer.parseInt(arr[i]);
        return res;
	}

    long[] longArr() throws IOException {
        String[] arr=nextLine().split("\\s");
		long[] res=new long[arr.length];
		for(int i=0;i<arr.length;i++) res[i]=Long.parseLong(arr[i]);
        return res;
    }

    double[] doubleArr() throws IOException {
		String[] arr=nextLine().split("\\s");
		double[] res=new double[arr.length];
		for(int i=0;i<arr.length;i++) res[i]=Double.parseDouble(arr[i]);
        return res;
    }

    String nextLine() throws IOException {
        return in.readLine();
    }
	
	InputStream stdin;
	OutputStream stdout;
	public problemB(){
		stdin=System.in;
		stdout=System.out;
	}
	
	public problemB(InputStream stdin, OutputStream stdout){
		this.stdin=stdin;
		this.stdout=stdout;
	}
}