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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _create_topk_unique(inputs, k):
 """Creates the top k values in sorted order with indices. Args: inputs: A tensor with rank of 2. [batch_size, original_size]. k: An integer, number of top elements to select. Returns: topk_r2: A tensor, the k largest elements. [batch_size, k]. topk_indices_r2: A tensor, indices of the top k values. [batch_size, k]. """

  height = inputs.shape[0]
  width = inputs.shape[1]
  neg_inf_r0 = tf.constant(-np.inf, dtype=tf.float32)
  ones = tf.ones([height, width], dtype=tf.float32)
  neg_inf_r2 = ones * neg_inf_r0
  inputs = tf.where(tf.is_nan(inputs), neg_inf_r2, inputs)

  # Select the current largest value k times and keep them in topk_r2. The
  # selected largest values are marked as the smallest value to avoid being
  # selected again.
  tmp = inputs
  topk_r2 = tf.zeros([height, k], dtype=tf.float32)
  for i in range(k):
    kth_order_statistic = tf.reduce_max(tmp, axis=1, keepdims=True)
    k_mask = tf.tile(tf.expand_dims(tf.equal(tf.range(k), tf.fill([k], i)), 0),
                     [height, 1])
    topk_r2 = tf.where(k_mask, tf.tile(kth_order_statistic, [1, k]), topk_r2)
    ge_r2 = tf.greater_equal(inputs, tf.tile(kth_order_statistic, [1, width]))
    tmp = tf.where(ge_r2, neg_inf_r2, inputs)

  log2_ceiling = int(math.ceil(math.log(float(int(width)), 2)))
  next_power_of_two = 1 << log2_ceiling
  count_mask = next_power_of_two - 1
  mask_r0 = tf.constant(count_mask)
  mask_r2 = tf.fill([height, k], mask_r0)
  topk_r2_s32 = tf.bitcast(topk_r2, tf.int32)
  topk_indices_r2 = tf.bitwise.bitwise_and(topk_r2_s32, mask_r2)
  return topk_r2, topk_indices_r2