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++11 (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <deque>
#include <algorithm>

using namespace std;

void printVector(deque<int> data) {
    auto iter = data.begin();
    cout << *iter++;
    for (; iter != data.end(); iter++) {
        cout << " " << *iter;
    }
    cout << endl;
}

int main(void) {
    int n;
    cin >> n;

    char command[20];
    int value;
    deque<int> data;

    for(int i = 0; i < n; i++) {
        scanf("%s%d", command, &value);
        if (command[0] == 'i') {
            data.push_front(value);
        }
        else if (command[0] == 'd' && command[6] == '\0') {
            deque<int>::iterator iter = find(data.begin(), data.end(), value);
            if (iter != data.end()) {
                data.erase(iter);
            }
        }
        else if (command[0] == 'd' && command[6] == 'F') {
            data.pop_front();
        }
        else if (command[0] == 'd' && command[6] == 'L') {
            data.pop_back();
        }
    }
    printVector(data);


    return 0;
}