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.*;
import java.lang.*;
import java.io.*;

	public class FastIO{
			
		BufferedReader br;
		StringTokenizer st;
		
		public FastIO(){ // constructor
			
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next(){
			
			while (st == null || !st.hasMoreElements()){
				try{
					st = new StringTokenizer(br.readLine());
				}
				
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt(){
			
			return Integer.parseInt(next());
		}
		
		long nextLong(){
			
			return Long.parseLong(next());
		}
		
		double nextDouble(){
			return Double.parseDouble(next());
		}

		String nextLine(){
			
			String str = "";
			try{
				
				str = br.readLine();
			}
			catch (IOException e){
				
				e.printStackTrace();
			}
			return str;
		}
		
		static int gcd(int a, int b) {
			
			if(b == 0) {
				return a;
			}
			
			return gcd(b, a % b);
		}
		
	public static void main(String[] args) throws IOException {
		
		FastIO io = new FastIO();
		
		// int n = io.nextInt(); -> for int
		// double d = io.nextDouble(); -> for double
		// String ch = io.next(); -> for one word
		// String str = io.nextLine(); -> for the entire line
		
		int n = io.nextInt();
		
		int arr[] = new int[n + 1];
		for(int i = 1; i <= n; i++) {
			arr[i] = io.nextInt();
		}
		
		int end = 0;
		int start = 0;
		
		for(int i = 1; i <= n; i++) {
			if(arr[i] != i) {
				end = arr[i];
				start = i;
				break;
			}
		}
		
		if(start == 0 && end == 0) {
			System.out.println(0 + " " + 0);
		}
		
		else {
			
			for(int i = start + 1; i <= end; i++) {
				if(arr[i] != arr[i - 1] - 1) {
					System.out.println(0 + " " + 0);
					return;
				}
			}
			
			for(int i = end + 1; i <= n; i++) {
				if(arr[i] != i) {
					System.out.println(0 + " " + 0);
					return;
				}
			}
			
			System.out.println(start + " " + end);
		}
 	}
}