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 C++ (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
struct node{
  node *prev;
  int key;
  node *next;
};
node L;
node *nil=&L;

node* find(int x){
  node *i;
  for(i=nil;i->next->key!=x;i=i->next);
  return i->next;
}
void del(int x){
  node *a;
  if(x==-1)a=nil->next; 
  else if(x==-2)a=nil->prev;
  else a=find(x);
  a->prev->next=a->next;
  a->next->prev=a->prev;
  free(a);
}
void in(int x){
  node *a;
  a=(node *)malloc(sizeof(node));
  a->next=nil->next;
  nil->next->prev=a;
  nil->next=a;
  a->prev=nil;
  a->key=x;
}

int main(){
  int n;
  string s;
  L.prev=nil,L.next=nil;
  scanf("%d\n",&n);
  while(n--){
    getline(cin,s);
    if(s[0]=='i')in(s[7]-'0');
    else{
      if(s[6]=='F')del(-1);
      else if(s[6]=='L')del(-2);
      else del(s[7]-'0');
    } 
  }
  node *i;
  for(i=nil;i->next->next!=nil;i=i->next)cout<<i->next->key<<' ';
  cout<<i->next->key<<endl;
  return 0;
}