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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class B {

	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		String line = in.readLine();
		int seq[] = new int[n];
		StringTokenizer tokenizer = new StringTokenizer(line, " ");
		for (int i = 0; i < n; i++) {
			seq[i] = Integer.parseInt(tokenizer.nextToken());
		}
		int e = 1;
		boolean done = false;
		boolean correct = true;
		int b = -1,f = -1;
		for (int i = 0; i < seq.length; i++) {
			if(seq[i] != e) {
				if(b == -1 ) {
					b = i;
				} 
				if(!done){
					f = i;
				}
				if(done) {
					correct = false;
					break;
				}
			} else {
				if(i+1<seq.length && seq[i+1] == seq[i]-1)
					f = i;
				else if(b != -1) {
					done = true;
				}
			}
			e ++;
		}
		
		if(correct && b != -1) {
			e = 1;
			for (int i = 0; i < b; i++) {
				if(seq[i] != e) {
					correct = false;
					break;
				}
				e++;
			}
			for (int i = f; i >= b && correct; i--) {
				if(seq[i] != e) {
					correct = false;
				}
				e++;
			}
			for (int i = f+1; i < seq.length && correct; i++) {
				if(seq[i] != e) {
					correct = false;
				}
				e++;
			}
		}
		if(!correct) {
			System.out.println("0 0");
		} else {
			System.out.printf("%d %d%n", b+1 , f+1);
		}
//		for (int i = 0; i < seq.length; i++) {
//			if(seq[i] != e) {
//				if(done) {
//					correct = false;
//					break;
//				}
//				b = i; f = seq[i] - 1;
//				for (int j = seq[i]-1; j >= i; j--) {
//					if(seq[j] != e) {
//						correct = false;
//					}
//				}
//			}
//			e++;
//		}
	}

}
