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 <stdio.h>
#include <string.h>
int B[99000];
char C[100];
int main(void){
	int i,j,n,k;
	int F,L;
	char str[1000];

	fgets(str,sizeof(str),stdin);
	sscanf(str,"%d",&n); F=-1; L=0;
	for(i=0;i<n;i++){
		fgets(str,sizeof(str),stdin);
		sscanf(str,"%s %d",&C,&B[i]);
		if(strncmp(C,"deleteFirst",11)==0){
			B[F]=-1; B[i]=-1;
		}
		else if(strncmp(C,"deleteLast",10)==0){
			B[L]=-1; B[i]=-1;
		}
		else if(strncmp(C,"delete",6)==0){
			for(j=F;j>=L;j--){
				if(B[j]==B[i]){
					B[j]=-1;
					break;
				}
			}
			B[i]=-1; 
		}
		for(j=0;j<=i;j++){
			if(B[j]!=-1){
				L=j;
				break;
			}
		}
		for(j=i;j>=0;j--){
			if(B[j]!=-1){
				F=j;
				break;
			}
		}
		/*printf("L %d F %d\n",L,F);
		for(k=0;k<=i;k++){
			printf("%d ",B[k]);
		}
		printf("\n");*/
	}
	j=0;
	for(i=n-1;i>=0;i--){
		if(B[i]!=-1){
			if(j!=0 && i!=n-1) printf(" ");
			printf("%d",B[i]);
			j++;
		}
	}
	printf("\n");
	return 0;
}