For a dynamic array $A = \\{a_0, a_1, ...\\}$ of integers, perform a sequence of the following operations:

* push($d$, $x$): Add element $x$ at the begining of $A$, if $d = 0$. Add element $x$ at the end of $A$, if $d = 1$.
* randomAccess($p$): Print element $a_p$.
* pop($d$): Delete the first element of $A$, if $d = 0$. Delete the last element of $A$, if $d = 1$.



$A$ is a 0-origin array and it is empty in the initial state.

Constraints

* $1 \leq q \leq 400,000$
* $0 \leq p < $ the size of $A$
* $-1,000,000,000 \leq x \leq 1,000,000,000$

Input

The input is given in the following format.


$q$
$query_1$
$query_2$
:
$query_q$


Each query $query_i$ is given by


0 $d$ $x$


or


1 $p$


or


2 $d$


where the first digits 0, 1 and 2 represent push, randomAccess and pop operations respectively.

randomAccess and pop operations will not be given for an empty array.

Output

For each randomAccess, print $a_p$ in a line.

Example

Input

11
0 0 1
0 0 2
0 1 3
1 0
1 1
1 2
2 0
2 1
0 0 4
1 0
1 1


Output

2
1
3
4
1
import java.util.*;
import java.io.*;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		ArrayDeque<Integer> deque = new ArrayDeque<>();

		int loop = in.nextInt();
		for(int i = 0; i < loop; i++) {
			int odr = Integer.parseInt(in.next()), wild = Integer.parseInt(in.next());

			switch(odr) {
				case 0:
					if(wild == 0 ) deque.addFirst(Integer.parseInt(in.next()));
					else deque.addLast(Integer.parseInt(in.next()));
					break;

				case 1:
					Iterator<Integer> itr = deque.iterator();
					for(int j = 0; j < wild; j++)itr.next();
					out.println(itr.next());
					break;

				case 2:
					if(wild == 0)deque.removeFirst();
					else deque.removeLast();
					break;
			}
		}
		out.flush();
		in.close();
	}

}

