The cardinality of the set of all linear order types over $\omega$ is $2^{\aleph_0}+\aleph_1$ in ZF+AD?

In ZFC, cardinality of set of linear orders over $$\omega$$ is $$2^{\aleph_0}$$. By the argument given by here, we can prove (without the choice) the number of linear orders over $$\omega$$ is at least $$2^{\aleph_0}$$. In addition, we can prove if $$A$$ is a set of countable linear order-types then $$|A|\ge \aleph_1$$ in ZF.

Therefore we can prove $$|A|\ge 2^{\aleph_0}$$ and $$|A|\ge\aleph_1$$ in ZF. If we assume the choice then we can prove $$|A|\le 2^{\aleph_0}$$. However, if we assume the AD (in fact, it is enough that assuming $$\aleph_1$$ and $$2^{\aleph_0}$$ are incomparable) then $$|A|>2^{\aleph_0}$$.

My question is : if we assume the AD (or, every subset of reals is Lebesgue measurable) then $$|A|=2^{\aleph_0}+\aleph_1$$? If not, there are known results about the cardinality of set of countable linear-order types?

The set of countable linear-order types are defined as follows : Let $$C\subset \mathcal{P}( \omega\times \omega)$$ be a set of all linear order over $$\omega$$. Define $$\le_1\,\sim\, \le_2$$ iff $$(\omega,\le_1)$$ and $$(\omega,\le_2)$$ is isomorphic (as linearly ordered set.) Then $$\sim$$ is a equivalence relation over $$C$$, and $$C/\sim$$ is a set of all 'linear order-types' over a countable set.

Nice question!

To get us started, a simple variant of the argument at that link gives us that there is an injection from $\omega_1^{<\omega_1}$ into $C/\sim$: Given such a sequence $(\alpha_\iota\mid \iota<\beta)$, consider the ordered sum $$\underbrace{\alpha_0+(\omega^*+\omega)+\alpha_1+(\omega^*+\omega)+\dots}_\beta,$$ where the underbrace is simply my poor notation to indicate that the sum continues transfinitely to include all $\alpha_\iota$ in the sequence.

The set $\omega_1^{<\omega_1}$ of (well-ordered) countable sequences of countable ordinals is much larger than $\mathfrak c+\omega_1$ (see for instance Woodin's paper on The cardinals below $|[\omega_1]^{<\omega_1}|$).