A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$.  Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$.  The side length of this square can be expressed in the form $\frac{a\sqrt{b}-c}{d}$, where $b$ is not divisible by the square of any prime, and $a$, $c$, and $d$ share no common factors. Find $a+b+c+d$.
[asy]
unitsize(15mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;

real a=1; real b=2;
pair O=(0,0);
pair A=(-(sqrt(19)-2)/5,1);
pair B=((sqrt(19)-2)/5,1);
pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5);
pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5);
pair E=(-(sqrt(19)-2)/5,0);
path inner=Circle(O,a);
path outer=Circle(O,b);
draw(outer); draw(inner);
draw(A--B--C--D--cycle);

draw(O--D--E--cycle);

label("$A$",D,NW);
label("$E$",E,SW);
label("$O$",O,SE);
label("$s+1$",(D--E),W);
label("$\frac{s}{2}$",(E--O),S);

pair[] ps={A,B,C,D,E,O};
dot(ps);
[/asy] Let $s$ be the length of the side of the square. The circles have radii of $1$ and $2$. We can then draw the triangle shown in the figure above and write expressions for the sides of the triangle in terms of $s$. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can use the Pythagorean Theorem: \begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\
\frac14 s^2 + s^2 + 2s + 1 &= 4\\
\frac54 s^2 +2s - 3 &= 0\\
5s^2 + 8s - 12 &=0.
\end{align*}Finally, we can use the quadratic formula to solve for $s$: $$s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \frac{2\sqrt{19}-4}{5}.$$Thus, our answer is $2+19+4+5=\boxed{30}$.