quantum field theory - Scattering amplitude and LSZ formula

I'm arriving at a contradiction.

To calculate the scattering amplitude, one usually follows the prescription given by the Feynman rules that you only consider fully connected diagrams with the required number of incoming and outgoing external legs (See Peskin & Schroeder pg 111 where they say: Only fully connected diagrams contribute to the $T$ matrix).

By fully connected, one means that you consider only graphs from which you can get from one line to any other line (See page 3 of this document).

On the other hand, we have the LSZ formula, which says that the scattering amplitude is given by the residue (as the momenta go on-shell) of the corresponding correlation function. For example, in $\phi^4$ theory, \begin{align} &\mathcal{M}(p_a,p_b \to k_1, k_2) \delta^{(4)}(p_a + p_b-k_1 -k_2) \sim \nonumber\\ &\lim_{p_a^2,p_b^2,k_1^2,k_2^2 \to m^2} (p_a^2 - m^2)(p_b^2 - m^2)(k_1^2 - m^2)(k_2^2 - m^2)G(p_a,p_b,-k_1,-k_2). \end{align}

But these two prescriptions seems to give a contradiction. Consider in $\phi^4$ theory, the $\mathcal{M}(4 \to 4)$ scattering. We have this diagram (ok if someone could draw the diagram that'll be great),

\begin{align} \text{X} \text{X} \end{align}

which consists of two separate $2 \to 2$ scattering processes.

This diagram is not fully connected, so we should ignore it by the first prescription, yet, it does not evaluate to $0$ under the LSZ formula, so we should include it.

Physically it makes sense that the leading order contribution to a $4 \to 4$ process is given by two separate $2 \to 2$ ones, but the fully connected prescription misses that out.

So, is there a caveat to the fully connected rule of drawing Feynman diagrams, since I believe the LSZ formula is mathematically true and physically reasonable?