Let
p(x, y)=a_{0}+a_{1} x+a_{2} y+a_{3} x^{2}+a_{4} x y+a_{5} y^{2}+a_{6} x^{3}+a_{7} x^{2} y+a_{8} x y^{2}+a_{9} y^{3} .
Suppose that
\begin{aligned}
p(0,0) & =p(1,0)=p(-1,0)=p(0,1)=p(0,-1) \\
& =p(1,1)=p(1,-1)=p(2,2)=0
\end{aligned}
There is a point \left(\frac{a}{c}, \frac{b}{c}\right) for which p\left(\frac{a}{c}, \frac{b}{c}\right)=0 for all such polynomials, where a, b, and c are positive integers, a and c are relatively prime, and c>1. Find a+b+c.