Use scala fastparse to remove enclosing escaped quotes but preserving the others

Question

I want to use fastparse to transform the following string \"Escaped quote\"\"\" into Escaped quote\". I have the following code which almost works.

def escapedQuote[_: P]: P[Unit] = P("\"")
def unquotedColumn[_: P] = P(escapedQuote ~ ((!escapedQuote ~ AnyChar.!) | escapedQuote ~ escapedQuote.!).rep ~ escapedQuote ~ End)
val result = parse(input, unquotedColumn(_))

I get the result ArrayBuffer(E, s, c, a, p, e, d, , q, u, o, t, e, ") which is very close to what I want, however I want the result in a String.

However, when I try the following (adding .! after the rep),

def unquotedColumn[_: P] = P(escapedQuote ~ ((!escapedQuote ~ AnyChar.!) | escapedQuote ~ escapedQuote.!).rep.! ~ escapedQuote ~ End)

I get the result Escaped quote\"\". An extra escaped quote has appeared.

I want to use my unquotedColumn in other parsers.

Any ideas how to fix my code?