Point P lies on the diagonal A C of square A B C D with A P>C P. Let O_{1} and O_{2} be the circumcenters of triangles A B P and C D P, respectively. Given that A B=12 and \angle O_{1} P O_{2}=120^{\circ}, then A P=\sqrt{a}+\sqrt{b}, where a and b are positive integers. Find a+b.