Corrected grid table size and references links

					splitCells.append([colspanMarker])

				else:

					parts = cell.split('<br />')

					# Check for overly long lines and split them too

					new_parts = []

					for part in parts:

						if len(part) > MAX_CELL_WIDTH:

							# Simple splitting by space

							words = part.split(' ')

							current_line = []

							current_length = 0

							for word in words:

								if current_length + len(word) + 1 > MAX_CELL_WIDTH:

									new_parts.append(' '.join(current_line))

									current_line = [word]

									current_length = len(word)

								else:

									current_line.append(word)

									current_length += len(word) + 1

							if current_line:

								new_parts.append(' '.join(current_line))

						else:

							new_parts.append(part)

					parts = new_parts

					if len(parts) > 1:

						# Found line breaks in cell

						# Add "\" to each part except the last

						parts = [ p + '\\' if i < len(parts)-1 else p 

								for i, p in enumerate(parts) ]

						# BUT only if it was originally a <br /> break, or if we want to indicate wrap?

						# For Markdown grid tables, simple newlines are enough.

						# But standard Grid Tables don't support multiline cells natively without some trickery?

						# Actually, Pandoc grid tables just use newlines.

						# The original code added `\` which might be for a specific renderer?

						# Standard reStructuredText / Pandoc grid tables treat lines in the same cell as just continuation.

						# No backslash needed usually.

						pass

						# parts = [ p + '\\' if i < len(parts)-1 else p 

						# 		for i, p in enumerate(parts) ]

					splitCells.append(parts)

				maxLines = max(maxLines, len(parts))

			# If we found line breaks, create multiple content lines

			# Each row's height is determined by its own content (maxLines for that row)

			# All cells in a row must have the same number of lines for proper alignment

			if maxLines > 1:

				for line_idx in range(maxLines):

				# Find the last line index that has content in any cell

				lastContentLineIdx = -1

				for check_idx in range(maxLines):

					for cellParts in splitCells:

						if len(cellParts) > check_idx and cellParts[check_idx].strip() and cellParts[check_idx].strip() != colspanMarker:

							lastContentLineIdx = check_idx

							break

				# Only create lines up to and including the last line with content

				for line_idx in range(lastContentLineIdx + 1):

					newCells = []

					for cellParts in splitCells:

						if len(cellParts) == 1 and cellParts[0].strip() == colspanMarker:

			for i in range(len(lines)):

				line = lines[i]

				if (m := _definitionExpression.match(line)) is not None:

					lines[i] = f'- <a name="_ref_{m.group(1)}"[{m.group(1)}]">[{m.group(1)}]</a>{m.group(2)}'

					# Use HTML anchor for definitions with span wrapper: <span id="_ref_1"><a name="_ref_1">[1]</a></span>

					lines[i] = f'<span id="_ref_{m.group(1)}"><a name="_ref_{m.group(1)}">[{m.group(1)}]</a></span>{m.group(2)}'

			def _repl(m:re.Match) -> str|None:

				if m.group(1) == '"':