Inverse of time-ordered exponential

It is straightforward to show that
$$ left[T_leftarrowexpleft(-iint_0^tH(tau)dtauright)right]^{-1} = T_rightarrowexpleft(iint_0^tH(tau)dtauright), $$
where $T_leftarrow$ and $T_rightarrow$ impose time-ordering and anti-time-ordering, respectively.

I’m wondering what the inverse of
$$ left[T_leftarrowexpleft(-iint_0^tint_0^tau H(tau)H(tau’)dtau,dtau’right)right]^{-1} = ? $$
might be. Just choosing anti-time-ordering doesn’t work. Anti-time-ordering of the "first" time index ($tau$ in this case) also fails.