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.util.Scanner;
public class SpoiltPer {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int[] arNums = new int[sc.nextInt()];
		int nReverse, init, fin;
		boolean rev, cor;
		rev = cor = false;
		nReverse = init = fin = 0;
		for(int i=0; i<arNums.length; i++) {
			arNums[i] = sc.nextInt();
		}
		int i;
		for(i=0; i<arNums.length - 1; i++) {
			if(arNums[i] != i + 1) rev = true;
			if(rev) {
				if(arNums[i+1] == (arNums[i] - 1)) {
					if(init == 0) init = i + 1;
					else fin = i + 1;
					cor = true;
				}
				else {
					if(fin != 0) fin = i + 1;
					rev = cor = false;
					nReverse++;
				}
			}
		}
		if(cor) fin = i + 1;
		if(nReverse > 1) System.out.println(0 + " " + 0);
		else System.out.println(init + " " + fin);
	}
}
