如何从泛型函数中省略prop类型？

declare function func1<T extends {...}>(params: {age: number, something: T, ...more}): string

declare function func2<FunctionType extends (...args: any) => any>(cb: FunctionType): FunctionType
const newFunction = func2(func1)  // returns same type as func1
newFunction<{...}>({something: {...}, age: ...}) // is valid, but age should not be valid

在func2中，如何从FunctionType的第一个参数中删除属性age？
我试过：

declare function func2<FunctionType extends (...args: any) => any>(cb: FunctionType): 
(params: Omit<Parameters<FunctionType>[0], 'age'>) => ReturnType<FunctionType>

const newFunction = func2(func1)  // This returns the new function type without age, but also without generics
newFunction<{...}>({...})   // the generic is no longer valid, bit is should be

这样我就可以删除属性age，但是返回类型不再是泛型的。我怎样才能保持它的泛型并从第一个参数中省略age呢？