Construct a PDA for at least one $w_i$ $\neq$ $w_{i + 1}^r$? My assignment asks
Let $S =
\{
w_1\#w_2\# \dots \#w_k | k ≥ 2; (∀i ≤ k)w_i ∈ \{0, 1\}^\star
; (∃i &lt; k) w_i\neq w_{i + 1}^r$
, i.e., not every string $w_i$ is equal to the reversal of $w_{i+1}$. Note $S ⊆ \{0, 1, \#\}^\star$.

Show that S is a CFL by a PDA, P, that accepts it.

I have been stuck on this question for approximately 20+ hours. I tried to sketch the PDA about 20 pages but still fail some cases. My friend's suggestion as store the word $w_i$ then the next state is to compare $w_{i + 1}$ to $w_i$, however, what I'm struggling is how do we compare $w_{i + 2}$ to $w_{i + 1}$ after comparing $w_{i + 1}$ to $w_{i}$. This seems impossible since the stack has to be pop out an element in order to compare the previous word. Once it's empty, we can never fill up the stack with the same elements that already popped. 

Another way that I found is referring to this link:
https://cseweb.ucsd.edu/classes/su99/cse105/hw2sol.pdf.

On question 3, it has a construction for non-palindrome over $\{0,1\}^\star$. Which is similar to our case but we have a $\#$ sign between words. I.e., we have many different strings instead of this construction only need to handle one string and compare it's left and right. I have been stuck for a long time, I think I'm missing something here. Any suggestions?
Use nondeterminism to guess an index $i$ for which $w_i \neq w_{i+1}$. The machine reads its input until it nondeterministically chooses a $w_i$ and places it on the stack. Afterwards, it compares $w_{i+1}$ to the content on the stack symbol by symbol. If the comparison fails, the machine accepts.