Problem:
Doubly Linked List

Your task is to implement a double linked list.

Write a program which performs the following operations:

insert x: insert an element with key x into the front of the list.

delete x: delete the first element which has the key of x from the list. If there is not such element, you need not do anything.

deleteFirst: delete the first element from the list.

deleteLast: delete the last element from the list.

Input

The input is given in the following format:

n

command1

command2

...

commandn

In the first line, the number of operations n is given. In the following n lines, the above mentioned operations are given in the following format:

insert x

delete x

deleteFirst

deleteLast

Output

Print all the element (key) in the list after the given operations. Two consequtive keys should be separated by a single space.

Constraints

The number of operations ≤ 2,000,000

The number of delete operations ≤ 20

0 ≤ value of a key ≤ 109

The number of elements in the list does not exceed 106

For a delete, deleteFirst or deleteLast operation, there is at least one element in the list.

Sample Input 1

7
insert 5
insert 2
insert 3
insert 1
delete 3
insert 6
delete 5

Sample Output 1

6 1 2

Sample Input 2

9
insert 5
insert 2
insert 3
insert 1
delete 3
insert 6
delete 5
deleteFirst
deleteLast

Sample Output 2

1

Solution in Ruby (judge verdict: Accepted)

class Node
  def initialize(prev_node, next_node, value)
    @value = value
    @prev = prev_node
    @next = next_node
  end
  attr_accessor :next_node, :prev_node, :value
end

class LinkedList
  def initialize
    @first_node = Node.new(nil, nil, nil)
    @last_node = Node.new(nil, nil, nil)
    first_node.next_node = last_node
    last_node.prev_node = first_node
  end
  attr_accessor :first_node, :last_node

  def insert(x)
    new_node = Node.new(first_node, nil, x)
    tmp = first_node.next_node
    connect_node new_node, tmp
    connect_node first_node, new_node
  end

  def delete(x)
    current = first_node.next_node
    until current.next_node.nil?
      if current.value == x
        connect_node current.prev_node, current.next_node
        break
      end
      current = current.next_node
    end
  end

  def deleteFirst
    connect_node first_node, first_node.next_node.next_node
  end

  def deleteLast
    connect_node last_node.prev_node.prev_node, last_node
  end

  def print_all
    current = first_node
    array = []
    until current.next_node.next_node.nil?
      array << current.next_node.value
      current = current.next_node
    end
    puts array.join ' '
  end

  def connect_node(first, second)
    first.next_node = second
    second.prev_node = first
  end
end

list = LinkedList.new
gets.to_i.times do
  input = gets.split(' ')
  case input[0]
  when 'insert'
    list.insert input[1].to_i
  when 'delete'
    list.delete input[1].to_i
  when 'deleteFirst'
    list.deleteFirst
  when 'deleteLast'
    list.deleteLast
  end
end
list.print_all