>(patt?)<sup>-n</sup></em>.
That is, it matches at most <code>|n|</code>
occurrences of <code>patt</code>.
</p>

<p>
In particular, <code>patt^0</code> is equivalent to <em>patt*</em>,
<code>patt^1</code> is equivalent to <em>patt+</em>,
and <code>patt^-1</code> is equivalent to <em>patt?</em>
in the original PEG notation.
</p>

<p>
In all cases,
the resulting pattern is greedy with no backtracking
(also called a <em>possessive</em> repetition).
That is, it matches only the longest possible sequence
of matc