[xsl] escape quotation marks in regular expression

["Wolfhart Totschnig wolfhart.totschnig@xxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Dear list,

I am receiving strings that may or may not be wrapped in double 
quotation marks. I would like to remove the quotation marks at the start 
and end of the string, if there are. There may also be quotation marks 
in the middle of the string, which should not be removed. I thought to 
use the replace() function for this, like so:

B B B  select="replace(., '^\"?(.+)\"?$', '$1')"

But Saxon throws the following error:

B B B  FORX0002: Syntax error at char 3 in regular expression: Escape 
character '"' not allowed

So what do I have to put instead of "\""? That is, what is the 
correct way to escape the quotation marks in the regular expression?

(I am aware that I could do this without a regular expression, i.e., 
with substring() and string-length(), but I would like to know how to do 
it with replace() since that seems simpler.)

Thanks in advance for your help!

Wolfhart