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 CF56B {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] arr = new int[n];
        int start = 0;
        int startP = 0;
        boolean startFlag = true;
        int end = 0;
        int endP = 0;
        for(int i=0;i<n;i++){
            int a = scanner.nextInt();
            arr[i] = a;
            if(i > 0 && startFlag){
                if(a != arr[i-1] + 1){
                    startP = i;
                    start = a;
                    startFlag = false;
                }
            }
        }
        for(int i=n-1;i>=0;i--){
            int a = arr[i];
            if(startP>= 1 && a == arr[startP-1] + 1){
                endP = i;
                end = arr[i];
                break;
            }
        }

        if(n == 2 && arr[0] > arr[1]){
            System.out.println(1 + " "+ 2);
        }
        else{
            if(startP>0 && endP>0 && startP < endP){
                int [] newData;
                newData = Arrays.copyOfRange(arr, startP, endP);
                int [] newData1;
                newData1 = Arrays.copyOfRange(arr, endP+1, n);
                if(isArrayInSortedOrderDesc(newData , newData.length) && isArrayInSortedOrderAsc(newData1, newData1.length)){
                    System.out.println(end + " "+ start);
                }
                else {
                    System.out.println("0 0");
                }
            }else {
                System.out.println("0 0");
            }
        }
    }

    private static boolean isArrayInSortedOrderDesc(int[] a, int index) {
        if (index < 2) {
            return true;
        } else {
            return (a[index-1] < a[index - 2]) && isArrayInSortedOrderDesc(a, index - 1);
        }
    }

    private static boolean isArrayInSortedOrderAsc(int[] a, int index) {
        if (index < 2) {
            return true;
        } else {
            return (a[index-1] > a[index - 2]) && isArrayInSortedOrderAsc(a, index - 1);
        }
    }
}