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
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

//package Aslan.codeforces;

/**
 *
 * @author Aslan
 */
import java.util.*;
public class ProblemB {
    public static void main(String args[]){
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[1002];
        for (int i=1;i<=n;i++)
            a[i] = in.nextInt();
        int cur = 0;

        for(int i=1;i<=n;i++){
            if (i!=a[i]){
                cur = i;
                break;
            }
        }

        if (cur!=0) {
            int j=cur;
            boolean p = true;

            for(int i=a[cur];i>=cur;i--){
                if (a[j]!=i){ p = false;break;}
                j++;
            }
            if (p == false) {System.out.println(0 + " " + 0); System.exit(0);}
        }

        if (cur!=0){
            for(int i=a[cur]+1;i<=n;i++){
                if (i!=a[i]){System.out.println(0 + " " + 0); System.exit(0);}
            }
        }

        System.out.println(cur + " " + a[cur]);
    }
}
   