Let $\langle O_1\cdots O_n\rangle$ be a Euclidean CFT$_d$ correlation function. I know that we can analytically continue to Lorentzian signature as follows. Let $x_i = (\tau_i,\mathbf{x}_i)\in\mathbb{R}^d$ be the positions of the operators. We set $\tau_i=i(t_i-i\epsilon_i)$. Then if $\epsilon_i>\epsilon_j$ the operator $O_i$ will be placed to the left of $O_j$ in the Lorentzian correlator. More precisely the Euclidean correlator in this configuration gives the Lorentzian vacuum expectation value $$\langle\Omega|O_{i_1}\cdots O_{i_n}|\Omega\rangle,\quad \text{provided $\epsilon_{i_1}>\epsilon_{i_2}>\cdots >\epsilon_{i_n}$}.$$
From these individual Wightman functions we can as always construct time-ordered correlators using theta functions. For the two-point function this is trivial $$\langle\Omega|T\{O_1O_2\}|\Omega\rangle=\langle\Omega|O_1O_2|\Omega\rangle\theta(t_1-t_2)+\langle \Omega|O_2O_1|\Omega\rangle\theta(t_2-t_1),$$
but for the three-point function it already becomes more involved. We get six terms, one for each Wightman function, each of which with two theta functions.
My question here is: is there a more direct route from the Euclidean three-point correlator $\langle O_1O_2O_3\rangle$ to the time-ordered Lorentzian correlator $\langle\Omega|T\{O_1O_2O_3\}|\Omega\rangle$ that bypasses the need to construct the later from the individual correlators for each of the operator orderings?
