[TypeScript] Combine Template Literal Types usage, Extract and default Generic type

For example we have a type like this:

type Obj = {
  a: "a";
  a2: "a2";
  a3: "a3";
  b: "b";
  b1: "b1";
  b2: "b2";
};

We want to constrct a new type which porps starts with letter 'a':

type keyStartingWithA = {
  [K in Extract<keyof Obj, `a${string}`>]: Obj[K];
};

So why this works?

Extract: get shared key from 'keyof Obj' (which is "a" | "a1" | "a2" | "b" | "b1" | "b2")

`a${string}` work as wired card: 'a*', * should be type string

So if we change to:

type keyStartingWithA = {
  [K in Extract<keyof Obj, `a${number}`>]: Obj[K];
};

OK, now we get Obj back starting with A, if we only want values of it.

type ValuesOfKeysStartingWithA = {
  [K in Extract<keyof Obj, `a${string}`>]: Obj[K];
}[Extract<keyof Obj, `a${string}`>];

Then we can improve the type to make it more generic

type ValuesOfKeysStartingWithA<T> = {
  [K in Extract<keyof T, `a${string}`>]: T[K];
}[Extract<keyof T, `a${string}`>];

type NewUnion = ValuesOfKeysStartingWithA<Obj>

Finially, you can notice that 

Extract<keyof T, `a${string}`

repeated two times, we can improve those by using default type of Generic types

type ValuesOfKeysStartingWithA<
  T,
  _ExtractedKeys extends keyof T = Extract<keyof T, `a${string}`>
> = {
  [K in _ExtractedKeys]: T[K];
}[_ExtractedKeys];

Creates a default variable _ExtractedKeys

Assign it to 

Extract<keyof T, `a${string}`>

Need to use following to keep it working

extends keyof T