How many ways to read the palindromic word ROTATOR in this design? The problem is as follows:

The figure from below shows the word $\textrm{ROTATOR}$ arranged in a
peculiar way. How many ways can this word be read assuming the equal
least distance from one letter to another?.


The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{490 ways}\\
2.&\textrm{480 ways}\\
3.&\textrm{245 ways}\\
4.&\textrm{400 ways}\\
\end{array}$
I noticed that the word is palindromic, hence it can be read back and forth, thus this means that I should account for these possibilites.
In order to keep the right track for this purpose I used an auxiliary numbers atop the letters to account for these as shown in the diagram from below.

After doing all of that I reached the conclusion that:
$\textrm{ways}=(74+74+96)\times 2 =488$
But this doesn't appear in any of the alternatives. Did I mess up something or what? Can someone help me here? Please I require a step by step explanation as I feel lost if my method did worked out properly?
 A: Here is a slightly easier argument.
Let's just count how many ways there are to go from an R to an A. The R's on one side give this:
$$R^1 \quad R^1 \quad R^1\\
O^1 \quad O^2 \quad O^2 \quad O^1 \\
T^3 \quad T^4 \quad T^3 \\
A^3 \quad A^7 \quad A^7 \quad A^3$$
Starting at the R's on the other end gives this:
$$A^4 \quad A^7 \quad A^7 \quad A^4 \\
T^1 \quad T^3 \quad T^4 \quad T^3 \quad T^1 \\
O^1 \quad O^2 \quad O^2 \quad O^1 \\
R^1 \quad R^1 \quad R^1$$
Adding those together, there are $7$, $14$, $14$, and $7$ ways to go from any R to each of the central A's. Conversely there are the same number of ways to go from each of those A's back to any R. Combining any R-to-A path with any A-to-R path from the same A we get $7\cdot7+14\cdot14+14\cdot14+7\cdot7 = 490$.
A: As @Peter hinted in a comment, we have to count for paths going from left to right, right to left as well as left to central $A$ column and back to left and so on.
Notation : $^{m}T^{n}$ indicates $T$ was first reached in $n$ number of ways and later $m$ total number of ways. Read right number first, left number second always.
Left to right paths (unidirectional):
$$
R \quad R \quad R\\
O^{1} \quad O^{2} \quad O^{2} \quad O^{1} \\
T^{3} \quad T^{4} \quad T^{3} \\
A^{3} \quad A^{7} \quad A^{7} \quad A^{3} \\
T^{3} \quad T^{10} \quad T^{14} \quad T^{10} \quad T^{3} \\
O^{13} \quad O^{24} \quad O^{24} \quad O^{13} \\
R^{37} \quad R^{48} \quad R^{37}
$$
Left + Right $= 2(37+48+37)=2\cdot 122 = 244$
Left to center and back to left : (top to bottom to top)
$$
^{34}R^{0} \quad ^{48}R^{0} \quad ^{34}R^{0}\\
^{10}O^{\color{red}{1}} \quad ^{24}O^{\color{red}{2}} \quad ^{24}O^{\color{red}{2}} \quad ^{10}O^{\color{red}{1}} \\
^{10}T^{3} \quad ^{14}T^{4} \quad ^{10}T^{3} \\
A^{3} \quad\quad A^{7} \quad\quad A^{7} \quad\quad A^{3}
$$
Left to center, back to left : $34+48+34=116$
Right to center and back to right : (bottom to top to bottom)
$$
A^{4} \quad\quad A^{7} \quad\quad A^{7} \quad\quad A^{4} \\
^{4}T^{1} \quad ^{11}T^{3} \quad ^{14}T^{4} \quad ^{11}T^{3} \quad ^{4}T^{1} \\
^{15}O^{\color{red}{1}} \quad ^{25}O^{\color{red}{2}} \quad ^{25}O^{\color{red}{2}} \quad ^{15}O^{\color{red}{1}} \\
^{40}R^{0} \quad ^{50}R^{0} \quad ^{40}R^{0}
$$
Right to center, back to right : $40+50+40=130$
Total ways : $244+116+130=490$
A: Starting from the top A there are $1+2+1+3=7$ routes to an R. So there are $7^2$ ways through that A.
Starting from second A down there are $(1+3+3)\times2=14$ routes to an R. So there are $14^2$ ways through that A.
So the total number of ways is $7^2+7^2+14^2+14^2=490$.
