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.Arrays;
import java.util.Scanner;

public class Tsk3 {

    static void metod() throws Exception {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        if (n == 1) {
            System.out.println("0 0");
        } else if (n == 2) {
            if (a[0] < a[1]) {
                System.out.println("0 0");
            } else {
                System.out.println("1 2");
            }
        } else {
            int l = -1;
            if (a[0] > a[1]) {
                l = 0;
            } else {
                for (int i = 1; i < a.length - 1; i++) {
                    if (a[i - 1] < a[i] && a[i] > a[i + 1]) {
                        l = i;
                        break;
                    }
                }
            }
            int r = -1;
            if (a[a.length - 2] > a[a.length - 1]) {
                r = a.length - 1;
            } else {
                for (int i = 1; i < a.length - 1; i++) {
                    if (a[i - 1] > a[i] && a[i] < a[i + 1]) {
                        r = i;
                        break;
                    }
                }
            }
        
            if (l == -1 || r == -1) {
                System.out.println("0 0");
            } else {
                rev(a, l, r);
                
                if (f(a)) {
                    System.out.println((l + 1) + " " + (r + 1));
                } else {
                    System.out.println("0 0");
                }
            }
        }

    }

    static boolean f(int[] a) {     
        for (int i = 0; i < a.length-1; i++) {
            if (a[i] > a[i + 1]) {
                return false;
            }
        }
        return true;
    }

    static void rev(int[] a, int l, int r) {
        for (int i = l; i <= l + (r - l) / 2; i++) {
            int k = a[i];
            a[i] = a[r + l - i];
            a[r + l - i] = k;
        }
    }

    public static void main(String[] args) throws Exception {
        Tsk3.metod();
    }
}