Complete Attack on RLWE Key Exchange with reused keys, without signal leakage I am studying a research paper "Complete Attack on RLWE Key Exchange with reused keys, without signal leakage" . On page number 21 to 28, there is toy example explaining the scheme. 

I am unable to get the desired value of $ \mathcal B$ as described in column 3 of table 1 on page number 22. My calculation is as follows: 

$s_B=(2,3,0,0,-5,2,3,1), \qquad e_B=(0,0,0,0,0,0,0,1)$ 

$e_B \cdot s_B=(x^7)\cdot(2+3x-5x^4+2x^5+3x^6+x^7)$

$e_B \cdot s_B=-3+5x^3-2x^4-3x^5-x^6+2x^7$

So, coefficient of $k_B[n-1]=k_B[7]=2 \cdot e_B \cdot s_B=2 \cdot 2=4$.

Now, calculating $sk_B[n-1]=sk_B[7]=\mathsf{Mod}_2(k_B[n-1],w_B[n-1])$

$=\mathsf{Mod}_2(4,1)= \left( 4+1 \cdot \dfrac{257-1}{2}  \mod  257 \right) \mod  2=\left(4+128 \mod \ 257 \right) \mod  2=0$

Thus, $sk_A=sk_B$ and value of oracle $\mathcal B$ must be $1$ (as keys get matched) but it is given $0$ (in column 3 of table 1 on page number 22).
There is a condition that is not considered, that is, the value after modulo 257 should be in $\mathbb Z_q$. 
When $q = 257,    \mathbb Z_q = \{ -128, ... , 128 \}$, so,  $(4+128)\mod 257$ should be $-125$ rather than $132$ .
And $-125 \mod 2 = 1$.
Thus, $sk_a \neq sk_b$  and the output of  oracle $\mathcal B$ is $0$.