3 ms·
It's repetitive - when you're checking e.g. for (ph|f)u(c|k|ck|q), there's no need to check cyb[e3]r(ph|f)u(c|k|ck|q) as well...
by vbar 15y ago
It's repetitive - when you're checking e.g. for (ph|f)u(c|k|ck|q), there's no need to check cyb[e3]r(ph|f)u(c|k|ck|q) as well...