return node.elemCount;
      }

      @Override
      long treeAggregate(@CheckForNull AvlNode<?> root) {
        return (root == null) ? 0 : root.totalCount;
      }
    },
    DISTINCT {
      @Override
      int nodeAggregate(AvlNode<?> node) {
        return 1;
      }

      @Override
      long treeAggregate(@CheckForNull AvlNode<?> root) {

                                .map(Key::toString)
                                .map(String::trim)
                                .sorted()
                                .distinct()
                                .collect(Collectors.joining("\n - ", " - ", "")));
            }
        }

        private <Q> Supplier<Q> getListSupplier(Key<Q> key) {

}

var (
	sqlLexer = lexer.Must(lexer.Regexp(`(\s+)` +
		`|(?P<Timeword>(?i)\b(?:YEAR|MONTH|DAY|HOUR|MINUTE|SECOND|TIMEZONE_HOUR|TIMEZONE_MINUTE)\b)` +

    return ObjectArrays.toArrayImpl(this, a);
  }

  @Override
  public Spliterator<E> spliterator() {
    return Spliterators.spliterator(this, Spliterator.ORDERED | Spliterator.DISTINCT);
  }

  @Override
  public void clear() {
    if (needsAllocArrays()) {
      return;
    }
    this.firstEntry = ENDPOINT;
    this.lastEntry = ENDPOINT;

    if (needsAllocArrays()) {
      return Spliterators.spliterator(new Object[0], Spliterator.DISTINCT | Spliterator.ORDERED);
    }
    Set<E> delegate = delegateOrNull();
    return (delegate != null)
        ? delegate.spliterator()
        : Spliterators.spliterator(
            requireElements(), 0, size, Spliterator.DISTINCT | Spliterator.ORDERED);
  }

  @Override
  public void forEach(Consumer<? super E> action) {

  /**
   * Performs a right rotation of {@code array} of "distance" places, so that the first element is
   * moved to index "distance", and the element at index {@code i} ends up at index {@code (distance
   * + i) mod array.length}. This is equivalent to {@code Collections.rotate(Booleans.asList(array),
   * distance)}, but is somewhat faster.
   *
   * <p>The provided "distance" may be negative, which will rotate left.
   *

  /**
   * Performs a right rotation of {@code array} of "distance" places, so that the first element is
   * moved to index "distance", and the element at index {@code i} ends up at index {@code (distance
   * + i) mod array.length}. This is equivalent to {@code Collections.rotate(Booleans.asList(array),
   * distance)}, but is somewhat faster.
   *
   * <p>The provided "distance" may be negative, which will rotate left.
   *

 *       hand, is multiple-use, and does implement {@link Iterable}.
 *   <li>Streams offer many features not found here, including {@code min/max}, {@code distinct},
 *       {@code reduce}, {@code sorted}, the very powerful {@code collect}, and built-in support for
 *       parallelizing stream operations.

      } else {
        return contains(element) ? unfiltered.remove(element, occurrences) : 0;
      }
    }
  }

  /**
   * Returns the expected number of distinct elements given the specified elements. The number of
   * distinct elements is only computed if {@code elements} is an instance of {@code Multiset};
   * otherwise the default value of 11 is returned.
   */

      } else {
        return contains(element) ? unfiltered.remove(element, occurrences) : 0;
      }
    }
  }

  /**
   * Returns the expected number of distinct elements given the specified elements. The number of
   * distinct elements is only computed if {@code elements} is an instance of {@code Multiset};
   * otherwise the default value of 11 is returned.
   */