* were inserted into the builder, equivalently to {@code LinkedHashMap}. For example, in the
   * above example, {@code WORD_TO_INT.entrySet()} is guaranteed to iterate over the entries in the
   * order {@code "one"=1, "two"=2, "three"=3}, and {@code keySet()} and {@code values()} respect
   * the same order. If you want a different order, consider using {@link ImmutableSortedMap} to

    @Override
    public Iterable<Entry<E>> order(List<Entry<E>> insertionOrder) {
      // We mimic the order from gen.
      Map<E, Entry<E>> map = new LinkedHashMap<>();
      for (Entry<E> entry : insertionOrder) {
        map.put(entry.getElement(), entry);
      }

      Set<E> seen = new HashSet<>();
      List<Entry<E>> order = new ArrayList<>();
      for (E e : gen.order(new ArrayList<E>(map.keySet()))) {

    }

    @Override
    public E[] createArray(int length) {
      return gen.createArray(length);
    }

    @Override
    public Iterable<E> order(List<E> insertionOrder) {
      return gen.order(insertionOrder);
    }
  }

  private static Set<Feature<?>> computeReserializedCollectionFeatures(Set<Feature<?>> features) {
    Set<Feature<?>> derivedFeatures = new HashSet<>(features);

		{
			input:          `comment="Hi this is my comment ="`,
			keys:           []string{"comment"},
			expectedFields: map[string]struct{}{`comment="Hi this is my comment ="`: {}},
		},
		// Keys and input order of k=v is same.
		{
			input: `connection_string="host=localhost port=2832" comment="really long comment"`,
			keys:  []string{"connection_string", "comment"},
			expectedFields: map[string]struct{}{

On Windows, [Current] has been made considerably faster when
the current user is joined to a slow domain, which is the
usual case for many corporate users. The new implementation
performance is now in the order of milliseconds, compared to
the previous implementation which could take several seconds,

  }

  /**
   * Returns an immutable list containing the given elements, in order.
   *
   * @throws NullPointerException if any element is null
   */
  public static <E> ImmutableList<E> of(E e1, E e2) {
    return construct(e1, e2);
  }

  /**
   * Returns an immutable list containing the given elements, in order.
   *
   * @throws NullPointerException if any element is null
   */

      }

      @Override
      public List<String> order(List<String> insertionOrder) {
        List<String> order = Lists.newArrayList();
        for (String s : insertionOrder) {
          int index = order.indexOf(s);
          if (index == -1) {
            order.add(s);
          } else {
            order.add(index, s);
          }
        }
        return order;
      }
    };
  }

  public void testCreate() {

    return new ImmutableLongArray(new long[] {e0});
  }

  /** Returns an immutable array containing the given values, in order. */
  public static ImmutableLongArray of(long e0, long e1) {
    return new ImmutableLongArray(new long[] {e0, e1});
  }

  /** Returns an immutable array containing the given values, in order. */
  public static ImmutableLongArray of(long e0, long e1, long e2) {

    return new ImmutableIntArray(new int[] {e0});
  }

  /** Returns an immutable array containing the given values, in order. */
  public static ImmutableIntArray of(int e0, int e1) {
    return new ImmutableIntArray(new int[] {e0, e1});
  }

  /** Returns an immutable array containing the given values, in order. */
  public static ImmutableIntArray of(int e0, int e1, int e2) {
    return new ImmutableIntArray(new int[] {e0, e1, e2});

     *
     * @since 3.5.0
     */
    List<MavenProject> getAllProjects();

    /**
     * Gets all projects in their intended build order, i.e. after topologically sorting the projects according to their
     * interdependencies.
     *
     * @return The projects in the build order, never {@code null}.
     */
    List<MavenProject> getSortedProjects();

    /**