Find the minimum value of \(n\) where \(n\) is an integer such that the number of trailing zeroes of \[(n+2)! \times (n + 0)! \times (n + 1)! \times (n + 4)!\] is divisible by \(2015\)

(\(201500000\) has \(5\) trailing zeroes while \(2000015000\) has \(3\))

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