Efficiently combine dataframes on 2nd level index

I have two dataframes looking like

import pandas as pd

df1 = pd.DataFrame([2.1,4.2,6.3,8.4,10.5], index=[2,4,6,8,10])
df1.index.name = 't'

df2 = pd.DataFrame(index=pd.MultiIndex.from_tuples([('A','a',1),('A','a',4),
                           ('A','b',5),('A','b',6),('B','c',7),
                           ('B','c',9),('B','d',10),('B','d',11),
                          ], names=('big', 'small', 't')))

I am searching for an efficient way to combine them such that I get

                 0
big small t       
A   a     1    NaN
          2    2.1
          4    4.2
    b     5    NaN
          6    6.3
B   c     7    NaN
          8    8.4
          9    NaN
    d     10  10.5
          11   NaN

I.e. I want to get the index levels 0 and 1 of df2 as index levels 0 and 1 in df1.

Of course a loop over the dataframe would work as well, though not feasible for large dataframes.

EDIT: It appears from comments below that I should add, the indices big and small should be inferred on t in df1 based on the ordering of t.

Solution

Assuming that you want the unknown index levels to be inferred based on the ordering of 't', we can use an other merge, sort the values and then re-create the MultiIndex using ffill logic (need a Series for this).

res = (df2.reset_index()
          .merge(df1, on='t', how='outer')
          .set_index(df2.index.names)
          .sort_index(level='t'))

res.index = pd.MultiIndex.from_arrays(
                [pd.Series(res.index.get_level_values(i)).ffill()
                 for i in range(res.index.nlevels)],
                names=res.index.names)

print(res)

                 0
big small t       
A   a     1    NaN
          2    2.1
          4    4.2
    b     5    NaN
          6    6.3
B   c     7    NaN
          8    8.4
          9    NaN
    d     10  10.5
          11   NaN