Product of treating 1,3‐dibromocyclobutane with alcoholic KOH at elevated temperatures Passage-II
  
  
  
  $$
\ce{
A -&gt;[\text{1.}~\ce{NH2-}(\pu{2 mol})][\text{2.}~\ce{CH3-I}(\pu{2 mol})]
B -&gt;[\ce{HgSO4}, \ce{H2SO4}][\ce{H2O}]
C -&gt;[\text{dil.}~\ce{OH-}][Δ]
D [Mixture]
}
$$
  
  31. "A" gives
  
  A) Lucas test
  B) Silver mirror test
  C) White precipitate with ammoniacal $\ce{AgNO3}$
  D) Yellow precipitate with ammoniacal cuprous chloride
  
  32. How many isomeric products are existed in the mixture "D"?
  
  A) 2
  B) 4
  C) 6
  D) 8 


What would be 'A' in the reaction? Any help would be deeply appreciated.

I know that alc. KOH does elimination, but I was confused whether it would eliminate one bromine each from two rings and create a bond between both the rings or it would simply eliminate both bromines from same ring and create two conjugated double bonds (cyclobut-1,3-diene).

EDIT: There is one more possibility: both the bromines get eliminated and a diagonal bond is formed in the ring (cyclobutane with any two diagonal carbons joined).
The two equivalents $\ce{HBr}$ are eliminated from two different molecules, one from the starting material and one from an intermediate. In the first step, $\ce{HBr}$ is eliminated under cleavage of the cyclobutane ring, which is favored by the release of ring strain. Elimination of $\ce{HBr}$ from 1-bromo-1,3-butadiene finally yields but-1-en-3-yne.