12 If $x^{5}+3 x^{3}+1=a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+\cdots+a_{5}(x-1)^{5}$ holds for any real number $x$, then the value of $a_{3}$ is $\qquad$. (Answer with a number)
12. In the expansion of $x^{5}+3 x^{3}+1=[(x-1)+1]^{5}+3[(x-1)+1]^{3}+1$, the coefficient of the $(x-1)^{3}$ term is $\mathrm{C}_{5}^{2}+3=13$, so $a_{3}=13$.