code: def bpopmin(self, timeout=0):
          
          res = self.database.bzpopmin(self.key, timeout)
          if res is not None:
              return (res[1], res[2])
docs: Atomically remove the lowest-scoring item from the set, blocking until
          an item becomes available or timeout is reached (0 for no timeout,
          default).
  
          Returns a 2-tuple of (item, score).