StackOverflowError for regexp with unescaped "@" symbol.

[RayzRazko]

Got java.lang.StackOverflowError if I try to generate random string with regexp that contains unescaped "@" symbol:

Generex rnd = new Generex("[a-z]@mail.com"); 
System.out.println(rnd.random());
//specifying the boundaries doesn't help 
Generex rnd = new Generex("[a-z]@mail.com"); 
System.out.println(rnd.random(20, 50));

while next line is totally ok for Java Pattern:
Pattern.compile("[a-z]@mail.com").matcher("m@mail.com").matches()

Generex works fine with next patterns:"[a-z]\\@mail.com", "[a-z][@]mail.com".

Stack:

java.lang.StackOverflowError
    at java.util.TimSort.countRunAndMakeAscending(TimSort.java:355)
    at java.util.TimSort.sort(TimSort.java:220)
    at java.util.Arrays.sort(Arrays.java:1438)
    at dk.brics.automaton.State.getSortedTransitionArray(Unknown Source)
    at dk.brics.automaton.State.getSortedTransitions(Unknown Source)
    at com.mifmif.common.regex.Generex.prepareRandom(Generex.java:282)
    at com.mifmif.common.regex.Generex.prepareRandom(Generex.java:303)
    at com.mifmif.common.regex.Generex.prepareRandom(Generex.java:303)
    at com.mifmif.common.regex.Generex.prepareRandom(Generex.java:303)
    at com.mifmif.common.regex.Generex.prepareRandom(Generex.java:303)
.....

[mifmif]

it seems that the @ is interpreted as a meta char , i will fix this issue.

[mifmif]

you can now use the feature discussed in issue #22 to avoid interpreting Metacharacter like @