Beautiful Attributes

{399!a(mod401)400!b(mod401)\large{\begin{cases} 399! \equiv a \pmod {401} \\ 400! \equiv b \pmod {401} \end{cases}}

If a,ba,b are the smallest possible positive integers satisfying the above modular equalities, find the value of a+b401\displaystyle \dfrac{a+b}{401} correct upto three places of decimals.

Note: You may use the fact that 401 is a prime number.

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