Let $\alpha,\beta,\gamma,\delta\in\mathbb{Z}$ and let $A(\alpha,\beta),\ B(1,0),\ C(\gamma,\delta),\ D(1,2)$ be the vertices of a parallelogram $ABCD$. If $AB=\sqrt{10}$ and the points $A$ and $C$ lie on the line $3y=2x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to: