how to split string characters into rows by order of appearance for each id?

Suppose, that I have a table composed by an id column and by a string represented by a set of characters:

with fake_data(id, ex_character) as (
    select * from values
        ('A', 'T70891'),
        ('B', 'RT9811111')
)

I would like to split string characters into rows for each different id value, so expected output should look like this:

# id   value

# A     T
# A     7
# A     0
# A     8
# A     9
# A     1
# B     R
# B     T
# B     9
# B     8
# B     1
# B     1
# B     1
# B     1
# B     1

So far I've tried something similar to:

select fd.id, f.value 

from fake_data fd, 
lateral split_to_table(regexp_replace(trim(fd.ex_character), '.', './/0', 2), ',') f

But this approach hasn't worked as it retrieves pivoted column unordered.

Do you know how may I retrieve expected data in an ordered way?

if you swap the split to be:

lateral split_to_table(regexp_replace(trim(fd.ex_character), '.', '.\
lateral split_to_table(regexp_replace(trim(fd.ex_character), '.', '.\\0', 2), '.') f
', 2), '.') f

it will happen correctly. to order thing you want to use f.index

thus:

with fake_data(id, ex_character) as (
    select * from values
        ('A', 'T70891'),
        ('B', 'RT9811111')
)
select 
    fd.id
    ,f.value
    ,f.index
from fake_data as fd,
lateral split_to_table(regexp_replace(trim(fd.ex_character), '.', '.\
with fake_data(id, ex_character) as (
select * from values
('A', 'T70891'),
('B', 'RT9811111')
)
select 
fd.id
,f.value
,f.index
from fake_data as fd,
lateral split_to_table(regexp_replace(trim(fd.ex_character), '.', '.\\0', 2), '.') f
order by fd.id, f.index
', 2), '.') f
order by fd.id, f.index

gives:

so the selecting of f.index is not needed, I just included it, so you could see what was available.