Three noncollinear points and a line $\ell$ are given in the plane. Suppose no two of the points lie on a line parallel to $\ell$ (or $\ell$ itself). There are exactly $n$ lines $m$ perpendicular to $\ell$ with the following property: the three circles with centers at the given points and tangent to line $m$ all concur at some point. Find all possible values of $n$.

Enter all possible values of $n,$ separated by commas.
The condition for line $m$ is that each of the three points lies at an equal distance from the line as from some fixed point; in other words, the line is the directrix of a parabola containing the three points, and the fixed point is the focus.

[asy]
unitsize (1 cm);

real func (real x) {
  return(x^2/4);
}

real a, b, c;
pair A, B, C, F;

a = -2; b = 1.5; c = 3;
A = (a,func(a));
B = (b,func(b));
C = (c,func(c));
F = (0,1);

draw(graph(func,-4,5),red);
draw(Circle(A,abs(A - F)));
draw(Circle(B,abs(B - F)));
draw(Circle(C,abs(C - F)));
draw((-4,-1)--(6,-1));
draw(A--F,dashed);
draw(B--F,dashed);
draw(C--F,dashed);
draw(A--(a,-1),dashed);
draw(B--(b,-1),dashed);
draw(C--(c,-1),dashed);
draw((-3,-2)--(-3,5.5));

label("$m$", (6,-1), E);
label("$\ell$", (-3,5.5), N);

dot(A);
dot(B);
dot(C);
[/asy]

Three noncollinear points in the coordinate plane determine a quadratic polynomial in $x$ unless two of the points have the same $x$-coordinate. Therefore, given the direction of the directrix, three noncollinear points determine a parabola, unless two of the points lie on a line perpendicular to the directrix. This case is ruled out by the given condition, so the answer is $\boxed{1}$.