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{int n;struct node *np,*pp;};
struct node *fp=NULL,*ep,*n,*nfp,*nep;
int N,t,a;
char cmd[15];

void del(struct node*dp)
{
	struct node *nnp=dp->np;
	struct node *npp=dp->pp;
	npp->np=nnp;
	nnp->pp=npp;
	free(dp);
}

int main()
{
	scanf("%d\n",&N);
	for(;N--;)
	{
		scanf("%s",cmd);
		if(cmd[0]=='i')
		{
			scanf("%d\n",&t);
			struct node *tp=(struct node*)malloc(sizeof(struct node));
			tp->n=t;
			if(fp==NULL)
			{
				tp->np=tp->pp=tp;
				fp=tp;
				ep=tp;
				tp->np=fp;
			}
			else
			{
				tp->np=fp;
				fp->pp=ep->np=tp;
				tp->pp=ep;
				fp=tp;
			}
		}
		else
		{
			if(cmd[6]=='F')
			{
				nfp=fp;
				fp=fp->np;
				del(nfp);
			}
			else if(cmd[6]=='L')
			{
				nep=ep;
				ep=ep->pp;
				del(nep);
			}
			else
			{
				scanf("%d\n",&t);a=0;
				for(n=fp;n->n!=t&&(!a||n!=fp);)
					n=n->np,a++;
				if(n->n!=t)continue;
				if(n==fp)
					fp=fp->np;
				if(n==ep)
					ep=ep->pp;
				del(n);
			}
		}
	}
	for(n=fp,t=1;t||n!=fp;n=n->np)
		t=!printf("%d%c",n->n,n->np==fp?'\n':' ');
	return 0;
}