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 $a+b+c$.