type parameter variance in generic call signature is incorrectly bivariant

まだ誰も着手していません。

### 🔎 Search Terms

unsound generic function assignability type parameter variance generic call signature bivariant

### 🕗 Version & Regression Information

- This is the behavior in every version I tried, and I reviewed the FAQ for entries about **generics and variance**.

### ⏯ Playground Link

https://www.typescriptlang.org/play/?ts=6.0.3#code/C4TwDgpgBAygFgewK4BsAmAhCBJAdgNwEMAnAS0N2AB4ARAPigF4oqAVOgCgA8AuKVgGQ0ANFBB9WASkZ1BNAFDyAxglwBnYFDUQVuNH3jJ0WPETIVqSXAGtcCAO64GzKN1EhJTBiGWqNUZGAAOSQAWwAjCGIDRFRMHAISckoqXDDI4mctHVU0eUCQiKiqK1sHJw4AFlEAcg1iGskAOmAEAFEuMFUISnIUDklFXX9SXEKMmKN40ySLVPSorO5PGVcuFvbO7t7CfslRLkHhzVGAVRs7R0m4k0TzFNLLpyYoUfGo+TOL8o464AbavVGoogA

### 💻 Code

```ts
type ShouldBeInvariant = (x: T&D, y: T)=>T&D

const second: ShouldBeInvariant = (x, y) => y
const outNumber: ShouldBeInvariant = second
outNumber(4, 'str').toExponential()

const inNumber: ShouldBeInvariant = (x) => (x.toExponential(), x)
const inUnknown: ShouldBeInvariant = inNumber
inUnknown('str', 'str')

```

### 🙁 Actual behavior

The code type checks, while it shouldn't because that is unsound, using `ShouldBeInvariant` covariantly or contravariantly both causes runtime errors.

### 🙂 Expected behavior

`ShouldBeInvariant` should be invariant against its parameter. Both assigns:
```ts
const outNumber: ShouldBeInvariant = second
```
and
```ts
const inUnknown: ShouldBeInvariant = inNumber
```
should report errors

### Additional information about the issue

_No response_

リンクされた TypeScript Playground の例から始め、ShouldBeInvariant を含む unsound な代入を再現してください。ジェネリックな呼び出しシグネチャの代入可能性と分散チェックを追跡し、両方の代入でエラーが報告され、無効な呼び出しが型チェックを通らなくなったことを示す回帰テストを追加してください。

索引モデルが issue の本文から書いたものです。