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
$$\sum\limits_{m=1}^{12}mm!=\sum_{m=1}^{12}(m+1-1)m! = \sum_{m=1}^{12}(m+1)!-m! = (12+1)!-1$$
