The greedy regex (cops) This is the cops' post. The robbers' is here.

You make a program and a regex.

For most inputs, one of the following happens:


the regex doesn't match, and the program returns a falsey value
the regex doesn't match, and the program returns a truthy value
the regex matches, and the program returns a truthy value


But for one (or more) input, the regex matches, but the program returns a falsy value.

You should try to keep this input hidden. Robbers will try to find an input where the regex matches, but the program returns a falsy value. 

Post your program and your regex, as well as your language name, byte count of the program, regex flavor, and byte count of the regex.

If a robber finds an input where the regex matches, but the program returns a falsy value, even if it wasn't the intended input your program is considered cracked, and you edit in a link to the robber's crack.

If nobody has cracked your program &amp; regex after a week, then edit in that your submission is safe and is therefore not eligible for cracking. Also edit in the intended input. The shortest (program bytes + regex bytes) non-cracked submission wins.

Example

Let's say your regex is (all of the flavors):

/^[0-9]$/


And your program is (JS):

x=&gt;Boolean(parseInt(x))


Then a valid crack would be 0, because the regex matches 0 but 0 is falsey.

Delimiters don't count in the byte count.

I can't enforce the banning of hash functions with rules, but please don't use SHA/RSA or anything that is unfair to the robbers
Python, cracked
Regex: (?!.)
Program: from operator import not_
This is a little easier than my previous one.