@{void} {insertAfter(Item e, Item after_item);} 
		{If the reference count of \verb+store+ is greater
			   than 1, then other collection objects are also 
			   using
			   it as their implementation.  Rather than changing
			   its contents, a copy will be made ($O(n)$), 
			   and the reference
			   counts will be updated.  Finally, 
			   insert $e$ after the first occurrence of 
			   \verb+after_item+.  If there is no such 
			   occurrence, a NULL pointer is passed to
			   the {\Container}'s 
			   {\em insertAfter(Item,ContainerNode*)} method. 
			   Complexity depends on the storage medium. }@

@{void} {insertBefore(Item e, Item before_item);} 
		{ Similar to {\em insertAfter()} except that $e$ is 
		  inserted before the first occurrence of 
		  \verb+before_item+. }@

@{void} {prepend(Item e);} {After considering reference counts as above, 
			   call \rar{store}{prepend(e)}, which should
			   insert $e$ at the beginning.  Note that this
			   usually has the same result as calling 
			   {\em insert}.}@

@{Item} {element(int i);} 
			{ Return the $i$th element, or the last element if
			  $i > \mbox{this->size()}$  $O(n)$.  }@ 

@{SequenceBase<Item,Impl>} {subsequence(int i, int j);} 
			{ Construct a new sequence and initialize
			  it with $s[i] \ldots s[j]$, where $s[p]$ is the 
			  $p$th element of the current sequence. $O(n)$}@