A stricter return type for a function with an optional fallbackValue parameter

I wrote a function that allows safe execution of another function and returns a default value in case of an error.

function safeExecute<T>(operation: () => T, fallbackValue?: T | undefined): T | undefined { try { return operation(); } catch (error) { return fallbackValue; } }

The fallbackValue parameter is optional. Therefore, the value returned by the safeExecute function has a type T | undefined. That's actually the question. I want the function to return `T` if the fallbackValue is set, and T | undefined if it is not set. So

safeExecute(() => 42, 0) -> number safeExecute(()=> "foobar") -> string | undefined

How I can achieve this? I tried using generic type for fallbackValue but no luck.