Typing result of function which is influenced by optional parameter

Example: https://flow.org/try/#0PQKgBAAgZgNg9gdzCYAoMHMdQYzgOwGcAXMYgEzAF4wAeAFQD4AKABwEMAndgW0IC4wAb3YB+Qc2YBKaozD0pAGjAAjCa049B9MADJhYHIJKcAlvgDmYAL4yqcgG5xT5W7OHosXz3iKkLAKbEAPIqAFbUYNLuQp5gnEEArpz4BkZgAOQ47IkA1gCeGTae1p4+BCTxAYShETSBIeHS5X5ghBqRHNx8AHTsYKJgXbyEfdGCQqUYw70qzEI9iwk14cqLPe2ctqhTXtgU857s-NH2USKCGewAFnBhGbaKnmpRMxPpJuYWysdtxGaWNxnWLTLi8PobRIqEzMACMygATFIdjtkXtsOiwKhgMA2qYeKwYAFOGAAgAPXiEgJgAC0YAAYvAkOQ4NUwPg4KRCAFqQAiOCsYimAjsGCmYj5Xlgcxka7UmZBYlgdj4SjEOUJKBwBLS1Lq6kJYjJVIOUWJAK4CqkAXEAKUGgMCaqYz-L42Fg24X4cTyKTaPQGX6fSw2GKeQ3GjxYX4ZW0kDJPLDrZieghgAA+6eEjxKKMx6CAA

StackOverflow question: https://stackoverflow.com/questions/47024349/typing-result-of-optional-function-parameter-featuring-in-return-value

The problem I face is that I have a return value of a function depending on optional parameter which is of generic type. Simple example here:

```js

// simpler example - Flow does not see "optionality" in the parameter and therefore in the return value
const opted = (option?: T): T & { a: string } => {
return {
a: 'test',
...(option || {}),
}
}

// Flow error:
24: return { ^ object literal. This type is incompatible with the expected return type of
23: const opted = (option?: T): T & { a: string } => { ^ some incompatible instantiation of `T`
```

Flow seems to don't understand that `option` argument is optional therefore the result of the function is also "optional".

Any idea how to type such code?