What is $p+4 \pmod {13}$ $p=\sum_{m=1}^{12} (m \cdot m!)$, and the question is just the title.
I tried applying Wilson's theorem i.e $(p-1)!+1\equiv0 \pmod p$
but did not get much help. Thanks for helping

Hint:-
$$\displaystyle\sum_{m=1}^{12}m\cdot m!=\displaystyle\sum_{m=1}^{12}(m+1)!-\displaystyle\sum_{m=1}^{12}m!=13!-1$$
