Using $\Delta_{\mathrm{f}} H_{298}^{\circ}$ And $S_{298}^{\circ}$ Data Given Below Calculate the Standard Gibbs Energy Change

Using $\Delta_{\mathrm{f}} H_{298}^{\circ}$ and $S_{298}^{\circ}$ data given below:
$\begin{array}{|l|l|l|l|l|}\hline & \mathrm{CH}_{4}(\mathrm{~g}) & \mathrm{O}_{2}(\mathrm{~g}) & \mathrm{CO}_{2}(\mathrm{~g}) & \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\\hline \Delta_{\mathrm{f}} H_{298}^{\circ} / \mathrm{kJ} \mathrm{mol}^{-1} & -74.8 & 0 & -393.5 & -285.8 \\\hline S_{298}^{\circ} / \mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1} & 186.3 & 205.1 & 213.7 & 69.9 \\\hline\end{array}$ Calculate the standard Gibbs energy change, $\Delta_{r} G_{298}^{\circ}$ (kJ mol^-1) , for the following reaction:
CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (l)

A) - 818.
B) + 71.5.
C) + 963.
D) - 963.