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.*;

public class Main {
	static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	public static void main(String[] args) throws IOException {
		int n = input.nextInt();
		int []a = new int[n+1];
		int []b = new int[n+1];
		for (int i=1;i<=n;i++){
			a[i] = input.nextInt();
		}
		int k = 0, flag = 0, l = 0;
		for (int i=1;i<=n;i++){
			if (a[i] != i){
				l = i;
				k = a[i]-i;
				int m = a[i];
				for (int j=l;j<=l+k;j++){
					b[j] = m;
					m -= 1;
				}
				flag = 1;
				break;
			}
		}
		if (flag == 0){
			pr.println("0 0");
			pr.flush();
			return;
		}else{
			for (int i=l;i<=l+k;i++){
				if (a[i] == b[i]){
					continue;
				}else{
					pr.println("0 0");
					pr.flush();
					return;
				}
			}
			for (int i=l+k+1;i<=n;i++){
				if (a[i] == i){
					continue;
				}else{
					pr.println("0 0");
					pr.flush();
					return;
				}
			}
		}
		pr.print(l + " " + a[l]);
		pr.flush();
	}
	
	static class input{
		static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		static StreamTokenizer in = new StreamTokenizer(br);
		static String nextLine() throws IOException{
			return br.readLine();
		}
		static int nextInt() throws IOException{
			in.nextToken();
			return (int)in.nval;
		}
		static long nextLong() throws IOException{
			in.nextToken();
			return (long)in.nval;
		}
		static double nextDouble() throws IOException{
			in.nextToken();
			return in.nval;
		}
	}
}