The given equation is equivalent to $(\frac{1}{x}-13)^{10}=-1$. Hence, any root $\gamma$ must satisfy $|\frac{1}{\gamma}-13|=1$. It follows $$(\frac{1}{\gamma_1}-13)(\frac{1}{\overline{\gamma_1}}-13)＋\cdots+(\frac{1}{\gamma_5}-13)(\frac{1}{\overline{\gamma_5}}-13)=5$$ Expand the above equation: $$(\frac{1}{\gamma_1 \overline{\gamma_1}}+\cdots+\frac{1}{\gamma_5 \overline{\gamma_5}})-13\times(\frac{1}{\gamma_1} + \frac{1}{\overline{\gamma_1}}+\cdots+\frac{1}{\gamma_5} + \frac{1}{\overline{\gamma_5}})+5\times 169=5$$ From the Vieta's theorem, we have $$\frac{1}{\gamma_1} + \frac{1}{\overline{\gamma_1}}+\cdots+\frac{1}{\gamma_5} + \frac{1}{\overline{\gamma_5}}=10\times 13$$ $$\therefore\quad \frac{1}{\gamma_1 \overline{\gamma_1}}+\frac{1}{\gamma_2 \overline{\gamma_2}}+\cdots+\frac{1}{\gamma_5 \overline{\gamma_5}} = 5-5\times 169 + 13\times 10\times 13=\boxed{850}$$