Unexpected distributive behaviour on type alias

### 🔎 Search Terms

distributive conditional types, naked type parameter, type alias, unions

### 🕗 Version & Regression Information

- This changed between versions 3.8.3 and 3.9.7.
- This is the behavior in every version I tried, and I reviewed the FAQ for entries about distributive conditional types, naked type parameter, type alias, unions.

### ⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.9.3#code/C4TwDgpgBAaghgGwK7QLxQIxQD5QEwBQBokUAKhALZhToDa8yEAulBAB7AQB2AJgM5Q6AS24AzCACdYzAlCgB+WG048BUJNwDW3APYB3bnPmKhMADSxEKWSagAuKNwgA3KccfO3koiWgAlCH4kBGBacioaAHoolUgAYzC6DEssXDxWXDo8VJx8WxMYgD0FIA

### 💻 Code

```ts
type Value = 1 | 2

type Result = [Value] extends [infer V]
? V extends unknown
? [V, Value]
: never
: never
```

### 🙁 Actual behavior

```ts
[1, 1] | [2, 2]
```

### 🙂 Expected behavior

```ts
[1, 1 | 2] | [2, 1 | 2]
```

### Additional information about the issue

It shouldn't work the way it does now, for several reasons:
1. Value is not naked type parameter.
2. Using brackets, distributive behaviour must be disabled.