An efficient (I guess) way to create a spiral in haskell as list of lists

spiral.hs

cons = (:)

baseSeed =
  [[1, 1, 1, 1, 1]
  ,[0, 0, 0, 0, 1]
  ,[1, 1, 1, 0, 1]
  ,[1, 0, 0, 0, 1]
  ,[1, 1, 1, 1, 1]]

baseSpiral = map (foldr (.) id) . map (map cons) $ baseSeed

spiralize :: Int -> [[Int]]
spiralize n = map ($ ([] :: [Int])) . map (spiralRow n) $ [1..n]
  where
    atEnd n i = i == 1 || i == n
    spiralRow size row
      | size < 5          = error "spiralize: size is too small"
      | size `rem` 2 == 0 = error "spiralize: size is even"
      | size == 5         = baseSpiral !! (row - 1)
      | otherwise         =
        if atEnd size row
        then foldr (.) id $ replicate size (cons 1)
        else
          if row == 2
          then (foldr (.) id $ replicate (size - 1) (cons 0)) . cons 1
          else
            if row == size - 1
            then cons 1 . (foldr (.) id $ replicate (size - 2) (cons 0)) . cons 1
            else cons 1 . cons corner . spiralRow (size - 4) (row - 2) . cons 0 . cons 1
            where
              corner = if row == 3 then 1 else 0

charify i = case i of
    1 -> '#'
    0 -> ' '
    _ -> error "Impossible!"

pretty = unlines . map (map charify)

main = putStrLn . pretty $ spiralize 205