Common initial sequence with nested struct

C++ has the "common initial sequence" rule for unions:

In a standard-layout union with an active member (11.5) of struct type T1, it is permitted to read a non-static data member m of another union member of struct type T2 provided m is part of the common initial sequence of T1 and T2; the behavior is as if the corresponding member of T1 were nominated.

How does this interact with nested structs? For instance, is it permissible to access u.s.s1.a1, u.s.s1.a2, u.s.s2.a1, u.s.s2.a2 if u.b is active? Or is the fact that S::s1 and B::b1 have different types imply that S and B do not have a common initial sequence?

struct A
{
  int a1, a2;
};

struct S
{
  A s1, s2;
};

struct B
{
  int b1, b2, b3, b4;
};

union U
{
  S s;
  B b;
};

/******************/
U u;
u.s.s1.a1 = 1;
u.b.b1 == 1;  // UB?
/******************/

The common initial sequence exceptions for inactive union member access is pretty specific. It effectively requires exactly identical non-static member definitions.

Specifically, [class.mem.general]/23 requires that, in declaration order, corresponding members have layout-compatible types.

That in turn is defined in [basic.types.general]/11 and can apply only if either both types are class types or neither is.

Since A and int are not both either class or non-class types, they are not layout-compatible and therefore the initial common sequence of S and B is empty.

So the access has UB.

It isn't even guaranteed for any but the first (indirect) int members that they have the same offset from their containing class. A conforming implementation would be allowed to add padding after the first int in A, but not in B.

Similarly, an implementation is allowed to choose alignof(A) != alignof(B).

Of course, neither of these is practical, so implementations don't do that.