The major product formed in the following reaction is
$C_6H_5CH=CH_2$ $\xrightarrow[(ii) H_3O^+]{(i) KMnO_4, KOH, \Delta}$ $\xrightarrow{(iii) Br_2/FeBr_3} \text{Product}$

• A
  $p$-bromophenylacetic acid
• B
  $o$-bromobenzoic acid
• C
  $m$-bromoacetophenone
• D
  $m$-bromobenzoic acid