If f(x)= {x+ a, x ≤ 0; |x-4|, x>0} and g(x)= {x+1, x<0; (x-4)^2+ b, x ≥ 0} are continuous on R, then (g o f)(2)+(f o g)(-2) is equal to:

If $\displaystyle f(x)=\left\{\begin{array}{ll}x+\mathrm{a}, & x \leq 0 \\ |x-4|, & x>0\end{array}\right.$ and $\displaystyle g(x)= \begin{cases}x+1, & x<0 \\ (x-4)^2+\mathrm{b}, & x \geqslant 0\end{cases}$ are continuous on $\displaystyle \mathbf{R}$, then $\displaystyle (g o f)(2)+(f o g)(-2)$ is equal to :