Lower Bound of countable order-types of linearly ordered sets On Page 44, Set theory, Jech(2006)

(Show that) there are at least $\mathfrak{c}$ countable order-types of linearly ordered sets. [For every sequence $a = \{ a_n : n \in N\}$ of natural numbers consider the order-type $\tau_a=a_0+\xi+a_1+\xi+a_2+\dots$, where $\xi$ is the order type of the integers. Show that if $a \ne b$, then $\tau_a \ne \tau_b$. ($\mathfrak{c}$ is the cardinality of $R$) ]

On Page 19, two sets have the same order type, if and only if they are isomorphic in the sense that there is a bijective function$f$, and both $f$ and $f^{-1}$ are order-preserving. So I speculate that the answer should equal the cardinality of the set of initial segments of the least uncountable ordinal, which is $\omega_1$. I must have misunderstood the problem somehow.
Even more confusing to me is the hint. 
Allow me to follow the convention in Jech's textbook:

  
*
  
*$1+\omega=\omega \ne \omega+1$
  
*$2·\omega=\omega \ne \omega·2 =\omega+\omega$
  

Since the order-type of integers is the sum of that of negative integers and that of natural number, which is $\omega·2$ . Then we have: $\tau_a = (a_0+\omega)+\omega+(a_1+\omega)+……=\omega^2$, which is a constant function, just contrary to injective function as shall be shown.
 A: I’ll use $\to$ to represent the order type of $\omega$ and $\leftarrow$ to represent that of $\omega^*$ and the negative integers, and I’ll use $\cdot$ to represent the order type of $1$. Then the order type of $3+\xi+2+\xi+0+\xi+1$ can be depicted thus:
$$\cdot\cdot\cdot\leftarrow\to\cdot\cdot\leftarrow\to\leftarrow\to\cdot$$
More generally, the order type $\tau_a=a_0+\xi+a_1+\xi+a_2+\xi+\ldots$ looks like this:
$$\underbrace{\cdot\cdot\ldots\cdot}_{a_0}\leftarrow\to\underbrace{\cdot\cdot\ldots\cdot}_{a_1}\leftarrow\to\underbrace{\cdot\cdot\ldots\cdot}_{a_2}\leftarrow\to\ldots$$
For each $k\in\Bbb N$ the last element of the $a_k$ section is identifiable by the fact that it’s the $(k+1)$-st element of the order that does not have an immediate successor; the first element of the $a_k$ section is identifiable by the fact that it’s the $(k+1)$-st element of the order that does not have an immediate predecessor. This means that each element that isn’t part of a $\xi$ block can be identified individually. Now use this idea to show that $\tau_a=\tau_b$ iff $a=b$.
Added: I thought that the symbolism of the arrows was self-explanatory, but if it isn’t, perhaps these diagrams will help:
$$\underset{\longrightarrow}{0,1,2,3,\ldots}$$
and
$$\underset{\longleftarrow}{\ldots,-4,-3,-2,-1}$$
