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.Scanner;

public class SpoiltPermutation {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = s.nextInt();
        }

        // two pointer approach
        int leftPtr = getLeftPointer(0, arr);
        int rightPtr = getRightPointer(arr.length -1 , arr);
        if(leftPtr == arr.length - 1 && rightPtr == 0) {
            System.out.println("0 0");
        } else if(decOrder(leftPtr, rightPtr, arr)) {
            if (leftPtr == 0 && rightPtr == arr.length-1) {
                System.out.println((leftPtr + 1) + " " + (rightPtr + 1));
            } else if(rightPtr + 1 < arr.length && arr[leftPtr] < arr[rightPtr+1] && leftPtr == 0) {
                System.out.println((leftPtr + 1) + " " + (rightPtr + 1));
            }
            else if (leftPtr-1 >= 0 && arr[rightPtr] > arr[leftPtr-1] && rightPtr == arr.length - 1) {
                System.out.println((leftPtr + 1) + " " + (rightPtr + 1));
            } else if (rightPtr + 1 < arr.length && leftPtr-1 >= 0 && arr[rightPtr] > arr[leftPtr-1] && arr[leftPtr] < arr[rightPtr+1]){
                System.out.println((leftPtr + 1) + " " + (rightPtr + 1));
            } else {
                System.out.println("0 0");
            }
        } else {
            System.out.println("0 0");
        }

    }

    static public int getLeftPointer(int start, int[] arr) {
        int j = start;
        while (j + 1 < arr.length && arr[j] < arr[j+1]) j++;
        return j;
    }

    static public int getRightPointer(int start, int[] arr) {
        int j = start;
        while (j - 1 >= 0 && arr[j] > arr[j-1]) j--;
        return j;
    }
    

    static public boolean decOrder(int start, int end, int[] arr) {
        for (int i = start; i < end; i++) {
            if (arr[i] < arr[i+1]) return false;
        }
        return true;
    }
}