How decode works in CCA1 scheme based on MP12 construction? In Section 6.3 from MP12 we have that $encode(m) = Sm$, for $S$ any basis of $\Lambda(G^t)$. Then I have:

$S = \begin{pmatrix}
1 &amp; 2 &amp; 4 &amp; 8 &amp; 16 &amp; 32 &amp; 64 &amp; 128 &amp; 256\\
2 &amp; 4 &amp; 8 &amp; 16 &amp; 32 &amp; 64 &amp; 128 &amp; 256 &amp; 0\\
4 &amp; 8 &amp; 16 &amp; 32 &amp; 64 &amp; 128 &amp; 256 &amp; 0 &amp; 0\\
8 &amp; 16 &amp; 32 &amp; 64 &amp; 128 &amp; 256 &amp; 0 &amp; 0 &amp; 0\\
16 &amp; 32 &amp; 64 &amp; 128 &amp; 256 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
32 &amp; 64 &amp; 128 &amp; 256 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
64 &amp; 128 &amp; 256 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
128 &amp; 256 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
256 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0
\end{pmatrix}$

I used very simple parameters: $q=512$, $k=9$, $n=1$, $m=18$. I set $H$ and $R$ to the identity matrix.This way I have that $A_u = [A' \mid -A' + G]$. 
Therefore, I have that $encode([1, 0, 0, 0, 0, 0, 0, 0, 0]) = [1, 2, 4, 8, 16, 32, 64, 128, 256] = G$. Fixing $s=1$, I computed $2s.A_u + e + encode(m) = [2.A' \mid -2A' +  2.G] + [0 \mid G] + e = [2.A' \mid -2A' + 3.G] + e \pmod{2q}$.

The inversion oracle returns $z=3$ and a big error $e'$ (but such that $e'.[R \mid I]^t = e.[R \mid I]^t \pmod{q}$). The decryption algorithm returns $m=[1, 1, 0, 0, 0, 0, 0, 0]$. 

In general I have that $m$ is being added to $2s$, so the decryption algorithm returns the binary decomposition of $DecimalRepresentation(m)+2s$, instead of only $m$.

In Lemma 6.2, we have that $v^t.[R \mid I]^t = 2(sG \pmod{q}) + encode(m) \pmod{2q}$. Therefore I reached this point successfully. Hence I guess my Decode algorithm is wrong. In order to decode, I am multiplying by $Inverse(S)$. Afterwards I divide the obtained vector by $q^{k-1}$.  

What am I doing wrong?
After some time I revisited the problem and found out that before decoding it is necessary to reduce modulo the fundamental region to find a coset, then we have that in the example I gave the internal value $2s+m$ is equal to $[1, 1, \dots]$, and after reduction it turns out to be equivalent to $[1, 0, \dots]$, which is $m$, as needed.