Triangle $\vartriangle ABC$ has sidelengths $AB = 10$, $AC = 14$, and, $BC = 16$. Circle $\omega_1$ is tangent to rays $\overrightarrow{AB}$, $\overrightarrow{AC}$ and passes through $B$. Circle $\omega_2$ is tangent to rays $\overrightarrow{AB}$, $\overrightarrow{AC}$ and passes through $C$. Let $\omega_1$, $\omega_2$ intersect at points $X, Y$ . The square of the perimeter of triangle $\vartriangle AXY$ is equal to $\frac{a+b\sqrt{c}}{d}$ , where $a, b, c$, and, $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a + b + c + d$.

Let's think step by step and output the final answer within \boxed{}.
user