human
Solve the following python programming problem given the following code:
def autoprec(ctx, f, maxprec = None, catch = (), verbose = False) : 
    '
        Return a wrapped copy of *f* that repeatedly evaluates *f*
        with increasing precision until the result converges to the
        full precision used at the point of the call.

        This heuristically protects against rounding errors, at the cost of
        roughly a 2x slowdown compared to manually setting the optimal
        precision. This method can, however, easily be fooled if the results
        from *f* depend "discontinuously" on the precision, for instance
        if catastrophic cancellation can occur. Therefore, :func:`~mpmath.autoprec`
        should be used judiciously.

        **Examples**

        Many functions are sensitive to perturbations of the input arguments.
        If the arguments are decimal numbers, they may have to be converted
        to binary at a much higher precision. If the amount of required
        extra precision is unknown, :func:`~mpmath.autoprec` is convenient::

            >>> from mpmath import *
            >>> mp.dps = 15
            >>> mp.pretty = True
            >>> besselj(5, 125 * 10**28)    # Exact input
            -8.03284785591801e-17
            >>> besselj(5, \'1.25e30\')   # Bad
            7.12954868316652e-16
            >>> autoprec(besselj)(5, \'1.25e30\')   # Good
            -8.03284785591801e-17

        The following fails to converge because `\\sin(\\pi) = 0` whereas all
        finite-precision approximations of `\\pi` give nonzero values::

            >>> autoprec(sin)(pi) # doctest: +IGNORE_EXCEPTION_DETAIL
            Traceback (most recent call last):
              ...
            NoConvergence: autoprec: prec increased to 2910 without convergence

        As the following example shows, :func:`~mpmath.autoprec` can protect against
        cancellation, but is fooled by too severe cancellation::

            >>> x = 1e-10
            >>> exp(x)-1.
expm1(x).
autoprec(lambda t: exp(t)-1)(x)
            1.00000008274037e-10
           ...

    
    

    Fix the buggy line: return ctx
assistant
#Here is Python program to fix the buggy line: return ctx
.

return def f_autoprec_wrapped(