Create Type based off of paramter type

Question

I have a generic function that returns another function.

When typing it, it might have required keys. Those required keys are optional for the first functional, but required for the second function IF they are not provided in the first. IF they are provided in the first function, they are then optional in the second (for overwriting)

Below is what I have tried, but haven't been able to figure it out.

function myFunc<T>(opts?: Partial<T>) {
  // this typing is wrong, but I don't know what it should be
  return (moreOpts: Omit<T, keyof typeof opts> & Partial<T>) => {
    // do stuff
  }
}

type TestType = {
  name: string;
  age: number;
}

const test1 = myFunc<TestType>({ name: 'John Doe' }) // no error
test1() // error -- needs age

const test2 = myFunc<TestType>() // no error
test2({ name: 'John Doe', age: 23 }) // no error

const test3 = myFunc<TestType>({ name: 'John Doe', age: 23 }); // no error
test3({ name: 'Jane Doe' }) // no error

const test4 = myFunc<TestType>({ name: 'John Doe' }); // no error
test4({ name: 'Jane Doe' }) // error -- needs age

const test5 = myFunc<TestType>({ name: 'John Doe', age: 23 }); // no error
test5() // no error