The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989 \ \text{cm}$. The length of each side of the triangle exceeds the length of each side of the square by $d \ \text{cm}$. The square has perimeter greater than 0. How many positive integers are NOT possible value for $d$?
$\text{(A)} \ 0 \qquad \text{(B)} \ 9 \qquad \text{(C)} \ 221 \qquad \text{(D)} \ 663 \qquad \text{(E)} \ \infty$
If t is the side of the triangle, and s is the side of the square, then
\[3t-4s=1989\]
\[t-s=d\]
Solving the first equation for t gives
\[t = \frac{4s+1989}{3}\]
Substituting into the second equation,
\[\frac{4s+1989}{3} - s = d\]
\[\frac{s+1989}{3} = d\]
\[s+1989 = 3d\]
If s = 0, d = 663. But s has to be greater than 0, so the first 663 positive integers aren't possible $\to\boxed{\textbf{(D)}}$