Modify TypeScript generic function parameters and return types [duplicate]

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Can I make all the arguments of a function readonly and convert that function to an async function?
For example:
declare function foo T (bar: T, baz: T[]): T;

declare function foo T (bar: T, baz: T[]): T;

Convert to
declare function foo T (bar: Readonly T , baz: readonly T[]): Promise T ;

declare function foo T (bar: Readonly T , baz: readonly T[]): Promise T ;

I've tried this, which works for non-generic functions:
type ReadonlyArrayItems T = { [P in keyof T]: Readonly T[P] };
type MakeFunctionAsync T = (...args: ReadonlyArrayItems Parameters T ) = Promise ReturnType T ;
type Foo = MakeFunctionAsync typeof foo ;

type ReadonlyArrayItems T = { [P in keyof T]: Readonly T[P] };
type MakeFunctionAsync T = (...args: ReadonlyArrayItems Parameters T ) = Promise ReturnType T ;
type Foo = MakeFunctionAsync typeof foo ;

However, the new function type Foo messes up the generic function by turning all the original generic type T into unknown.
Foo
T
unknown

P.S. Thanks for the answers.
However, is it possible to do this for all functions within an interface?
For example:
interface Methods {
foo T (bar: T, baz: T[]): T;
bar(baz: string): void;
}

interface Methods {
foo T (bar: T, baz: T[]): T;
bar(baz: string): void;
}

Become this
interface Methods {
foo T (bar: Readonly T , baz: readonly T[]): Promise T ;
bar(baz: string): Promise void ;
}

interface Methods {
foo T (bar: Readonly T , baz: readonly T[]): Promise T ;
bar(baz: string): Promise void ;
}

If add another type based on the above:
type MakeFunctionsAsync T = {
[functionName in keyof T]: MakeFunctionAsync T[functionName] ;
};

type MakeFunctionsAsync T = {
[functionName in keyof T]: MakeFunctionAsync T[functionName] ;
};

There doesn't seem to be a place to fill in the generic type parameter.
type MakeFunctionsAsync T, U = {
[functionName in keyof T]: MakeFunctionAsync T[functionName] U ;
};

type MakeFunctionsAsync T, U = {
[functionName in keyof T]: MakeFunctionAsync T[functionName] U ;
};

This won't work.

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