human
Solve the following python programming problem given the following code:
def parse_tokens(tokenrules, tokentexts, dictionarytype = dict) : 
    def inc(pos) : 
        return (pos + 1) 
def rule_at(pos) : 
        try : 
            return tokenrules [pos] 
except IndexError : 
            raise ParserError ('Expecting more tokens', pos) 
def text_at(pos) : 
        try : 
            return tokentexts [pos] 
except IndexError : 
            raise ParserError ('Expecting more tokens', pos) 
def skip_literal(pos, rulenr) : 
        if (rule_at (pos) == rulenr) : 
            return inc (pos) 
return pos 
def musthave_literal(pos, s) : 
        nextpos = skip_literal (pos, s) 
        if (nextpos == pos) : 
            raise ParserError (("Expecting '%s'" % s), pos) 
return nextpos 
def sequence(pos, * args) : 
        elements = [] 
        for element in args : 
            if callable (element) : 
                r = element (pos) 
                (v, pos) = r 
                elements.append (v) 
else : 
                pos = musthave_literal (pos, element) 
return (elements, pos) 
def zero_or_more(pos, func, nextelem) : 
        elements = [] 
        while True : 
            try : 
                (v, pos) = func (pos) 
                elements.append (v) 
except ParserError as e : 
                try : 
                    musthave_literal (pos, nextelem) 
except ParserError as e2 : 
                    e2.prevexc = e 
                    raise e2 
return (elements, pos) 
return (elements, pos) 
def anystring(pos) : 
        rulenr = rule_at (pos) 
        if (rulenr == RULE_UNQUOTEDSTRING) : 
            return unquotedstring (pos) 
else : 
            if (rulenr == RULE_QUOTEDSTRING) : 
                return quotedstring (pos) 
raise ParserError ('Expecting string', pos) 
def value(pos) : 
        rulenr = rule_at (pos) 
        if (rulenr == RULE_UNQUOTEDSTRING) : 
            return unquotedstring (pos) 
else : 
            if (rulenr == RULE_DICTIONARY) : 
                return dictionary (pos) 
else : 
                if (rulenr == RULE_QUOTEDSTRING) : 
           ...

    
    

    Fix the buggy line: return (rule_at, root)
assistant
#Here is Python code to fix the buggy line: return (rule_at, root)
.

return (success, root)