# your code goes here
import itertools
a=range(9)
print [list(map(lambda i_v:i_v[1],i_v)) for k, i_v in itertools.groupby(enumerate(a), lambda i_v:i_v[0]//3)]
IyB5b3VyIGNvZGUgZ29lcyBoZXJlCmltcG9ydCBpdGVydG9vbHMKYT1yYW5nZSg5KQpwcmludCBbbGlzdChtYXAobGFtYmRhIGlfdjppX3ZbMV0saV92KSkgZm9yIGssIGlfdiBpbiBpdGVydG9vbHMuZ3JvdXBieShlbnVtZXJhdGUoYSksIGxhbWJkYSBpX3Y6aV92WzBdLy8zKV0=
[[0, 1, 2], [3, 4, 5], [6, 7, 8]]