Does the reaction of 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one with NaOH involve ring contraction? What is the product and mechanism when 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one is reacted with
  $\ce{NaOH}$?
  
  


The only reagent used is $\ce{NaOH}$ so I suppose it abstracts the acidic hydrogen from the $\ce{-OH}$ group. A nucleophilic attack at the carbonyl carbon is also possible, as it could form a tropylium like cation. But I am not sure as to how a ring contraction occurs. The final answer given is 3-methyl-4-nitrobenzoic acid. How does the formation of an acid group and the ring contraction take place?
Drawing on the suggestion of @user55119 I propose the mechanism shown below. 

Whether the first intermediate bicyclic [4.1.0] heptanone arises from an intramolecular electrocyclic process or an intramolecular aldol is unclear but either seems possible. Once the highly strained bicyclic [4.1.0] heptanone intermediate is formed, the reaction proceeds with OH- attack on the carbonyl in a process analogous to a Favorskii mechanism here, with the second intermediate anion stabilised by the nitro group which also directs the direction of ring opening. Finally loss of OH- in a manner similar to base-catalysed aldol and driven by re-aromatisation gives the observed product.