Why is the C++ function parameter stored 20 bytes off of the rbp in x86-64 when the method body only has one 4 byte variable?

Question

Consider the following program, compiled using x86-64 GCC 12.2 with flags --std=c++17 -O0:

int square(int num, int num2) {
    int foo = 37;
    return num * num;
}

int main () {
    return square(10, 5);
}

The resulting assembly using godbolt is:

square(int, int):
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-20], edi
        mov     DWORD PTR [rbp-24], esi
        mov     DWORD PTR [rbp-4], 37
        mov     eax, DWORD PTR [rbp-20]
        imul    eax, eax
        pop     rbp
        ret
main:
        push    rbp
        mov     rbp, rsp
        mov     esi, 5
        mov     edi, 10
        call    square(int, int)
        nop
        pop     rbp
        ret

I read about shadow spaces and it appears that in x64 there must be at minimum 32 bytes allocated: "32 bytes above the return address which the called function owns" ...

With that said, how is the offset -20 determined for the parameter num? If there's 32 bytes from rbp, wouldn't that be -24?

I noticed even if you add more local variables, it'll remain -20 until it gets pushed over to -36, but I cannot understand why. Thanks!