Electronic – Charge distribution on switched capacitors

Computing the voltage and charge of these capacitors in series and parallel, I come up with a situation in which I couldn't explain why the circuit is behaving as it is. The circuit is the following:

^{simulate this circuit – Schematic created using CircuitLab}

Both switches SW1 and SW2 change their position at time \$t_1 = 1~ns\$.
Therefore, on the interval \$0 < t < (t_1 = 1~ns)\$, the capacitor \$C_1\$ is connected to the source \$I_1\$. At the end of this first interval:

$$V_1\left(t_1^-\right) = \frac{1}{C_1}\int_0^{t_1}{I_1 dt} = \frac{1}{1~pF}\int_0^{1~ns}{10^{-3}dt} = 1~V$$

and the charge at \$C_1\$ is \$Q|_{t_1^-} = 10^{-12}~C\$

Right after both switches SW1 and SW2 change the position and connect \$C_2\$ and \$C_3\$ to the circuit, the voltage at \$C_2\$ is negative and \$V_1\$ drops to \$0.5~V\$.

Making some numbers, I would have expected that the charge in \$C_1\$ would have been transfered to \$C_3\$, forcing a current from ground to \$V_1\$ node, and then charging negatively \$C_2\$. However, according to my calculations that would set \$V_{C_2}\left(t_1^+\right) = -0.333~V\$ since \$0.333\cdot10^{-12}~C\$ would have been transferred to \$C_3\$:
$$Q_{C_1} = Q|_{t_1^-}\frac{C_1}{C_1 + C_3} = 10^{-12} \frac{1~pF}{1~pF + 0.5~pF} = 0.666 \cdot 10^{-12}~C$$
$$Q_{C_2} = Q|_{t_1^-}\frac{C_3}{C_1 + C_3} = 10^{-12} \frac{0.5~pF}{1~pF + 0.5~pF} = 0.333 \cdot 10^{-12}~C$$

Obviously there is some mistake in all this reasoning. However, I can't spot it. Does anyone know what is exactly happening at \$t_1 = 1~ns\$ when both switches change the position and the charge at \$C_1\$ is redistributed, that would explain the values on the plot?