Two capacitors $C_1$ and $C_2$ (where $C_1 > C_2$) are charged to the same initial potential difference $\Delta V_i$. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity. The switches S1 and S2 are then closed. Find the final potential difference $\Delta V_f$ between $a$ and $b$ after the switches are closed as in figure down below.
The solution finds the initial charge $Q_i = Q_{1i}+Q_{2i} = C_1 \Delta V_i -C_2 \Delta V_i = (C_1 - C_2) \Delta V_i$. I don't understand why is there a negative sign for $Q_{2i}$. Can someone explain ?