*       size. In many cases {@link Ordering#leastOf} may work for your use case with significantly
 *       improved (and asymptotically superior) performance.
 *   <li>The retrieval operations {@link #peek}, {@link #peekFirst}, {@link #peekLast}, {@link
 *       #element}, and {@link #size} are constant-time.
 *   <li>The enqueuing and dequeuing operations ({@link #offer}, {@link #add}, and all the forms of

 *       size. In many cases {@link Ordering#leastOf} may work for your use case with significantly
 *       improved (and asymptotically superior) performance.
 *   <li>The retrieval operations {@link #peek}, {@link #peekFirst}, {@link #peekLast}, {@link
 *       #element}, and {@link #size} are constant-time.
 *   <li>The enqueuing and dequeuing operations ({@link #offer}, {@link #add}, and all the forms of

    public E getLast() {
      synchronized (mutex) {
        return delegate().getLast();
      }
    }

    @Override
    @CheckForNull
    public E peekFirst() {
      synchronized (mutex) {
        return delegate().peekFirst();
      }
    }

    @Override
    @CheckForNull
    public E peekLast() {
      synchronized (mutex) {
        return delegate().peekLast();

    public E getLast() {
      synchronized (mutex) {
        return delegate().getLast();
      }
    }

    @Override
    @CheckForNull
    public E peekFirst() {
      synchronized (mutex) {
        return delegate().peekFirst();
      }
    }

    @Override
    @CheckForNull
    public E peekLast() {
      synchronized (mutex) {
        return delegate().peekLast();

            }
        }

        // Visit the edges after sorting the components in consumer-first order
        while (!queue.isEmpty()) {
            ComponentState component = queue.peekFirst();
            if (component.getVisitState() == VisitState.NotSeen) {
                component.setVisitState(VisitState.Visiting);
                int pos = 0;
                for (NodeState node : component.getNodes()) {