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.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;


public class Main {

    static int [] nums;
    static int [] diff;
    public static void main(String[] args) {
        Scanner s = new Scanner(new InputStreamReader(System.in));
        int n = s.nextInt();
        nums = new int[n+1];
        diff = new int[n+1];
        
        for (int i = 1; i <= n; i++)
            nums[i] = s.nextInt();
        
        for (int i = 0; i < n; i++)
            diff[i] = Math.abs(nums[i] - nums[i+1]);
        diff[n]= Math.abs(n+1-nums[n]);
        
        //System.out.println(Arrays.toString(diff));
        int chg_num = 0;
        int chg1=-1,chg2=-1;
        int chg1_i = 0,chg2_i= 0 ;
        for (int i = 0; i <= n; i++) {
            if (diff[i] == 1) continue;
            chg_num++;
            if (chg_num > 2) break;
            if (chg1 == -1){
                chg1 = diff[i];
                chg1_i = i;
            }else if (chg2 == -1){
                chg2 = diff[i];
                chg2_i = i;             
            }
        }
        
        if (chg_num > 2 || chg_num == 0){
            System.out.println("0 0");
        }else if (chg_num ==1){
            if (chg1_i == 0)
                System.out.println("1 " + n);
            else
                System.out.println("0 0");
        }else{
            if (chg1 == chg2)
                System.out.println(chg1_i+1 + " " + chg2_i);
            else
                System.out.println("0 0");
        }
    }

}
