Let $N$ be the number of ways to place $4$ bishops on a $5 \times 5$ chessboard such that no $3$ are on the same diagonal. Find the remainder when $N$ is divided by $100$. (Note: the length of a diagonal on a $5 \times 5$ chessboard can be 2, 3, 4, or 5.)

Let's think step by step and output the final answer within \boxed{}.
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