Problem: 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

Solution (in JAVA):

Solution:
import java.util.*;
public class code2 {
    public static void main(String[]args){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int [] mas= new int [n+1];
        int x=0;
        int y=0;
        boolean p=true;
        for(int i=1;i<=n;i++){
            mas[i]=sc.nextInt();
            if(mas[i]!=i && p){
                x=i;
                y=mas[i];
                p=false;
            }
        }
        if(x==0){
            System.out.println(0);
            System.out.println(0);
        }
        else{
            int [] ans=new int [n+1];
        for(int i=1;i<x ;i++){
            ans[i]=i;
        }
        for(int i=0;i<=y-x;i++){
            ans[x+i]=y-i;
        }
        for(int i=y+1;i<n+1;i++){
            ans[i]=i;
        }
        if(com(mas, ans)){
            System.out.println(x);
            System.out.println(y);
        }
        else {
            System.out.println(0);
            System.out.println(0);
        }
        }
        
        
    }
    static boolean com(int [] mas, int [] ans){
        for(int i=1;i<mas.length;i++){
            if(mas[i]!=ans[i]) return false;
        }
        return true;
        
    }
}