Cities $A$, $B$, $C$, $D$, and $E$ are connected by roads $\widetilde{AB}$, $\widetilde{AD}$, $\widetilde{AE}$, $\widetilde{BC}$, $\widetilde{BD}$, $\widetilde{CD}$, and $\widetilde{DE}$. How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.) [asy]
size(5cm);

pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08);

dot (A);

dot (B);

dot (C);

dot (D);

dot (E);

label("$A$",A,S);

label("$B$",B,SE);

label("$C$",C,E);

label("$D$",D,N);

label("$E$",E,W);

guide squiggly(path g, real stepsize, real slope=45)

{

real len = arclength(g);

real step = len / round(len / stepsize);

guide squig;

for (real u = 0; u < len; u += step){

real a = arctime(g, u);

real b = arctime(g, u + step / 2);

pair p = point(g, a);

pair q = point(g, b);

pair np = unit( rotate(slope) * dir(g,a));

pair nq = unit( rotate(0 - slope) * dir(g,b));

squig = squig .. p{np} .. q{nq};

}

squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))};

return squig;

}

pen pp = defaultpen + 2.718;

draw(squiggly(A--B, 4.04, 30), pp);

draw(squiggly(A--D, 7.777, 20), pp);

draw(squiggly(A--E, 5.050, 15), pp);

draw(squiggly(B--C, 5.050, 15), pp);

draw(squiggly(B--D, 4.04, 20), pp);

draw(squiggly(C--D, 2.718, 20), pp);

draw(squiggly(D--E, 2.718, -60), pp);[/asy]
The presence of cities $C$ and $E$ is irrelevant to the problem, because upon entering either city, there is only one road going out. Therefore, we can remove those cities, and instead note that there are two roads connecting $A$ and $D,$ two roads connecting $B$ and $D,$ and one road connecting $A$ and $B.$ We can assume that the order in which each pair of roads is traversed does not matter, and then multiply by $2 \cdot 2 =4$ at the end.

Now, take cases on whether $B$ or $D$ is visited first:

Suppose $D$ is visited first. If the other road back to $A$ is then taken, then the only possibility is to travel to $B$ and then travel the two roads between $B$ and $D$ in either order. If, instead, one of the roads to $B$ is taken, then either $A, D, B$ must be visited in that order, or $D, A, B$ must be visited in that order. This gives $3$ possible routes in total.

Suppose $B$ is visited first. Then $D, A, D, B$ must be visited in that order, so there is only one possible route.

Putting the two cases together and multiplying by $4$ gives the answer, $4(1+3) = \boxed{16}.$