type parameter variance in generic call signature is incorrectly bivariant

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### 🔎 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_

Start with the linked TypeScript Playground example and reproduce the unsound assignments involving ShouldBeInvariant. Trace generic call-signature assignability and variance checking, then add a regression test showing that both assignments report errors and that the invalid calls no longer type-check.

Written by the indexing model from the issue text.