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: Accepted)

#include <iostream>
#include <string>
#include <list>

using namespace std;

void print_list(list<int> &V)
{
		for(list<int>::iterator it = V.begin();
			it != V.end();it ++)
		{
			if(it ==  V.begin()){
				cout << *it;
			}
			else{
				cout << " " << *it;
			}
		}
	cout << endl;
}
int main(void)
{	
	ios_base::sync_with_stdio(false);

	list<int> V;
	
	int n;
	cin >> n;

	for(int i = 0; i < n; i++){
		string s;
		cin >> s;
		
		int x;
		if(s == "insert"){
			cin >> x;
			//V.(V.begin(), x);
			V.push_front(x);
		}
		else if(s == "delete"){
			cin >> x;
			for(list<int>::iterator it = V.begin();
				it != V.end();it ++)
			{	
				if(*it == x){
					V.erase(it);
					break;
				}
			}
		}
		else if(s == "deleteFirst"){
			V.pop_front();
		}
		else{
			V.pop_back();
		}
	}
	print_list(V);
	
	return 0;
}