I have a trouble with the following sum $\sum_{i=0}^n\binom{n}{i}S(i,m)3^i$, where $S(i,m)$ is the Stirling number of the second kind (the number of all partitions of $i$ elements into $m$ nonempty sets).

Below it was obtained the sum $f(n,m)=\frac{1}{m!}\sum_{k=0}^m(-1)^{m-k}\binom{m}{k}(1+3k)^n$. Is there a further simplification of this expression?