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;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Map;
import java.util.Set;

public class Main { 
    
    // private static int dp[][][];
    // private static HashMap<String, Integer> dp = new HashMap<>();
    // add all number which has its any place as zero plus one no zero any number
	public static void main(String[] args) {
        Scanner ip = new Scanner(System.in);
        int n = ip.nextInt();
        int ar[] = new int[n];
        for (int i=0; i<n; i++) {
            ar[i] = ip.nextInt();
        }
        int start = -1;
        for (int i=0; i<n; i++) {
            if (ar[i] != i+1) {
                start = i;
                break;
            }
        }
        int end = -1;
        for (int i=1; i<n; i++) {
            if (ar[i] != i+1 && ar[i]<ar[i-1]) {
                end = i;
            }
        }

        if (start == -1 || end == -1) {
            System.out.println("0 0");
        } else {
            int s = start+1, e = end+1;
            while (start < end) {
                int tmp = ar[start];
                ar[start]= ar[end];
                ar[end] = tmp;
                start++;
                end--;            
            }
            boolean flag = false;
            for (int i=0; i<n-1; i++) {
                if (ar[i] >= ar[i+1]) {
                    flag = true;
                    break;
                }
            }
            if (flag) System.out.println("0 0");
            else System.out.println(s+" "+e);
        }

	}
    private static int LCS(String s1, String s2) {
        int dp[][] = new int[s1.length()+1][s2.length()+1];
        int max = 0;
        for (int i=0; i<=s1.length(); i++) {
            for (int j=0; j<=s2.length(); j++) {
                if (i == 0 || j == 0) {
                    dp[i][j] = 0;
                    continue;
                }
                if (s1.charAt(i-1) == s2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1]+1;
                    max = Math.max(max, dp[i][j]);
                } else {
                    dp[i][j] = 0;
                }
            }
        }

        return max;
    }
}