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 Main {
    static boolean isRe(int m[]){
        int n = m.length;
        int []b=new int [n];
        int count =n;
        for(int i=0;i<n;i++){
            if(i+1==m[i]){
                b[i]=2;
                count--;
            }
        }
        if(count==0){
            return false;
        }
        int f=0;int l=1000;
        for(int i=0;i<n;i++){
            if(b[i]!=2){
                f=i;
                i=1000;
            }
        }
        for(int i=n-1;i>=0;i--){
            if(b[i]!=2){
                l=i;
                i=-1000;
            }
        }       
        if(l<=f){
            l=0;f=0;
            return false;
        }
        boolean d=true;
        for(int i=f;i<l;i++){
            if(m[i]<m[i+1]){
                d=false;
            }
        }
        if(d){
            x=f+1;
            y=l+1;
        }
        return true;
    }
    static int x=0;
    static int y=0;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m[] = new int[n];
        for(int i=0;i<n;i++){
            m[i]=in.nextInt();
        }
        isRe(m);        
        System.out.print(x+" "+y);
    }

}