REGEXP_REPLACE removing leading and trailing quotes.

Hello guys,
I have an issue with my regular expression and I'm hoping that someone can help me with it?
Imagine the following sample_data:
{code}
WITH sample_data AS
SELECT '''[email protected]''' as e FROM dual UNION ALL
SELECT '''[email protected]' as e FROM dual UNION ALL
SELECT '[email protected]''' as e FROM dual UNION ALL
SELECT '''[email protected]''' as e FROM dual UNION ALL
SELECT '[email protected]''' as e FROM dual UNION ALL
SELECT '''[email protected]' as e FROM dual
SELECT e, REGEXP_REPLACE(e, '(^'')?(.*)(''$)', '\2')
FROM sample_data;
{code}
I would like to remove all leading and trailing quotes from the address email. I did the previous regexp but it does not returns the correct result. I don't know why, but for some reasons, when my email does not ends with a quote, the leading quote is not removed.
Can anyone help me with this?
Thanks,

Cant you just trim it?
WITH sample_data AS
SELECT '''[email protected]'''      as e FROM dual UNION ALL
SELECT '''[email protected]'        as e FROM dual UNION ALL
SELECT '[email protected]'''        as e FROM dual UNION ALL
SELECT '''[email protected]''' as e FROM dual UNION ALL
SELECT '[email protected]'''   as e FROM dual UNION ALL
SELECT '''[email protected]'   as e FROM dual
SELECT e, rtrim(ltrim(e, ''''), '''') new_e
FROM sample_data;

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