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.*;
// warm-up
public class SpoiltPermutation {

	static void solve(){
		Scanner sc = new Scanner(System.in);
		int noc = sc.nextInt(), i = 0, j = 0, n = noc, l = -1, r=noc;
		int[] a = new int[noc];
		boolean ok = true;
		while (noc-->0) a[i++] = sc.nextInt();
		i = 0;   while (++l<n  && a[l]==++i);
		i = n+1; while (--r>=0  && a[r]==--i);
		i=l; j=r;
		while (i<j) {
			int temp = a[i]; a[i] = a[j]; a[j] = temp;
			i++; j--;
		}
		for (i=0; i<n-1; i++) if (a[i]+1!=a[i+1]) { ok = false; break; }  
		System.out.println(ok && l<r ? (l+1)+" "+(r+1) : "0 0");
		sc.close();
	}

	public static void main(String args[]) {
		solve();
	}

}