Find the smallest $a\in\mathbb{N}$, so that $a\equiv3\cdot 5^4\cdot 11\cdot 13^3 \pmod 7$ The problem I am trying to solve is:

Find the smallest $a\in\mathbb{N}$, so that $a\equiv3\cdot 5^4\cdot 11\cdot 13^3 \pmod 7$.

I noticed that $3,5,11$ and $13$ are primes, but I've no idea how that is supposed to help me.
The solution is $4$, but why?

Note $13^{3} \equiv (-1)^{3} \equiv -1 \pmod 7$
Also, note that $5^{4} \equiv 25^{2} \equiv 4^{2} \equiv 2 \pmod 7$. 
Thus $3 \times 5^{4} \times 11 \times 13^{3} \equiv 3 \times 2 \times (-3) \times (-1) \equiv 18 \equiv 4 \pmod 7$. 
