smc

Divide the circle into four parts: the top semicircle by connecting E, F, and G($A$); the bottom sector ($B$), whose arc angle is $120^{\circ }$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$, giving a $30^{\circ }-60^{\circ }-90^{\circ }$ triangle; the triangle formed by the radii of $A$ and the chord ($C$); and the four parts which are the corners of a circle inscribed in a square ($D$). Then the area is $A+B-C+D$ (in $B-C$, we find the area of the bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter). The area of $A$ is $\frac{1}{2}\pi \cdot 2^2=2\pi$. The area of $B$ is $\frac{120^{\circ }}{360^{\circ }}\pi \cdot 2^2=\frac{4\pi }{3}$. For the area of $C$, the radius of $2$, and the distance of $1$ (the smaller semicircles' radius) to $BC$, creates two $30^{\circ }-60^{\circ }-90^{\circ }$ triangles, so $C$'s area is $2\cdot \frac{1}{2}\cdot 1\cdot \sqrt{3}=\sqrt{3}$. The area of $D$ is $4\cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi$. Hence, finding $A+B-C+D$, the desired area is $\frac{7\pi }{3}-\sqrt{3}+4$, so the answer is $7+3+3+4=\boxed{17}$.