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 <stdio.h>
#include <stdlib.h>
struct node_t {
	int key;
	struct node_t *next;
	struct node_t *prev;
};
typedef struct node_t Node;
static Node g_first, g_last;

void insert(int key) {
	Node *p;

	p = malloc(sizeof(Node));
	p->key = key;
	p->next = g_first.next;
	p->prev = &g_first;
	g_first.next->prev = p;
	g_first.next = p;
}

void delete_first(void) {
	Node *p;

	p = g_first.next;
	if (p != &g_last) {
		g_first.next = p->next;
		p->next->prev = &g_first;
		free(p);
	}
}

void delete_last(void) {
	Node *p;

	p = g_last.prev;
	if (p != &g_first) {
		g_last.prev = p->prev;
		p->prev->next = &g_last;
		free(p);
	}
}

void delete_key(int key) {
	Node *p;

	for (p = g_first.next; p != &g_last; p = p->next) {
		if (p->key == key) {
			p->prev->next = p->next;
			p->next->prev = p->prev;
			free(p);
			break;
		}
	}
}

void print_node(void) {
	Node *p;
	const char *fmt = "%d";

	for (p = g_first.next; p->key >= 0; p = p->next) {
		printf(fmt, p->key);
		fmt = " %d";
	}
	putchar('\n');
}

int main(void) {
	int n, key;
	char in[20];

	g_first.next = &g_last;
	g_last.prev = &g_first;
	g_first.key = g_last.key = -1;
	scanf("%d ", &n);
	while (n--) {
		scanf("%s%d", in, &key);
		if (in[0] == 'i') {
			insert(key);
		} else {
			if (in[6] == '\0') {
				delete_key(key);
			} else if (in[6] == 'F') {
				delete_first();
			} else {
				delete_last();
			}
		}
	}
	print_node();
	return 0;
}