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.*;


public class Main {


    public static void main(String[] args) {
        int n,l=0,r;
        int[] a = new int[1020];
        Scanner in = new Scanner(System.in);
        r = n = in.nextInt();
        a[0] = 0;
        for (int i = 1; i <= n; i++) {
            a[i] = in.nextInt();
        }

        while (a[l]==l&&l<=n) l ++;
        while (a[r]==r&&r>=1) r --;

        if (l==n+1&&r==0) {
            System.out.println("0 0");
            return;
        }
        for (int i = l; i < r; i ++) {
            if (a[i]<a[i+1]) {
                System.out.println("0 0");
                return;
            }
        }
        System.out.println(l+" "+r);
    }


}
 	      			    	  	 	 	 	  	