Two different color rooks are placed on a chessboard so they don't attack each other. What is a number of ways we can do that?

Of course we have $64$ choices for the first one and then $49$ choices for the second one. So we have $64\cdot 49$ ways to put them. But some of the configurations are essentially the same if we say rotate chessboard or reflect across some diagonal. So we should divide $64 \cdot 49$ with some number. But which? Is it $4$ since we have $4$ rotations which takes chessboard to it self? And then also by $4$ since we have $4$ reflections across the line? So the final result should be 
$$64\cdot 49\over 16$$
But I feel this is not correct, since if we have $7\times 7$ ''chessboard'' then we would get the result: $$49\cdot 36\over 16$$
which is not an integer.
 A: Hmm, I think I found a proper number. 
We can (essentialy) put white rook only on one of a marked cells in pictures (A) or (B):
(A):
\begin{array} {|r|r|r||r|r|r||r|r|}
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 \color{white}{*}& * &*& * & \color{white}{*} &\color{white}{*}  & \color{white}{*}  & \color{white}{*} \\
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\end{array}
We have $6$ choises for the white rook and of course $49$ for the black rook. So the answer in this case is $6\cdot 49=294$.
(B)
\begin{array} {|r|r|r||r|r|r||r|r|}
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 *&  &&  & \color{white}{*} &\color{white}{*}  & \color{white}{*}  & \color{white}{*} \\
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 & * & &  &  & &  & \\
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  &  &*&  &  & &  & \\
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 &  & & * &  & &  & \\
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&  & &  &  & &  & \\
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\end{array}
In this case we have $4$ choises for the white rook and $21$ for the black rook not on diagonal and $7$ if it is on diagonal. So the answer in this case is $4\cdot 28=112$.
Overall we have 406 ways to put two nonatacking rooks on a chessboard.
A: A rook that is not on a main diagonal can be taken to eight other squares by rotation and reflection.  A rook that is on the main diagonal can be taken to four other squares.  If the two rooks are not on the same main diagonal we should divide by $8$ because we can do four rotations and either a reflection or not.  If the two rooks are on the same diagonal we should only divide by $4$ because the reflection does not matter.  There are $2\cdot 8 \cdot 7=112$ positions with the two rooks on one diagonal, so there are $\frac {112}4=28$ that are different after considering rotation and reflection.  There are then $\frac 18(64\cdot 49-112)=378$ distinct positions with the rooks not on the same diagonal, for a total of $406$.
