lazy doesn't compile when used with default

[piotrmarkiewicz1]

here is example from tutorial, but with added field with default:

const baseCategorySchema = z.object({
  name: z.string(),
  val: z.number().default(0),
});

type Category = z.infer<typeof baseCategorySchema> & {
  subcategories: Category[];
};

const categorySchema: z.ZodType<Category> = baseCategorySchema.extend({
  subcategories: z.lazy(() => categorySchema.array()),
});

Compiler complains

Type 'ZodObject<extendShape<{ name: ZodString; val: ZodDefault<ZodNumber>; }, { subcategories: ZodLazy<ZodArray<ZodType<Category, ZodTypeDef, Category>, "many">>; }>, "strip", ZodTypeAny, { ...; }, { ...; }>' is not assignable to type 'ZodType<Category, ZodTypeDef, Category>'.
  Types of property '_input' are incompatible.
    Type '{ name: string; subcategories: Category[]; val?: number | undefined; }' is not assignable to type 'Category'.
      Type '{ name: string; subcategories: Category[]; val?: number | undefined; }' is not assignable to type '{ name: string; val: number; }'.
        Types of property 'val' are incompatible.
          Type 'number | undefined' is not assignable to type 'number'.
            Type 'undefined' is not assignable to type 'number'.ts(2322)

I believe this is valid case