Inspired by: this post

```
class Solution(object):
def flipLights(self, n, m):
m, n = min(3, m), min(3, n)
return 1 if n == 0 or m == 0 else self.flipLights(n - 1, m) + self.flipLights( n - 1, m - 1)
```

Operations: O(flip odds), E(flip evens), A(flip all), T(flip 3k + 1), N(flip nothing)

Relations:

O + O = N, E + E = N, A + A = N, T + T = N

O + E = A, O + A = E, E + A = O

Exclusive statuses :

n > 2:

① N

② O

③ E

④ A

⑤ T

⑥ O + T

⑦ E + T

⑧ A + T

n = 2 (remove all T related statuses):

① N

② O

③ E

④ A

n = 1(remove all T, E, A related statuses):

① N

② O

Steps needed to all status( always can plus 2 * k)

① can only be achieved by 0, 2 steps

②，③，④ can be achieved by either 1 or 2 steps

⑤ can only be achieved by 1 steps

⑥，⑦，⑧ can only be achieved by 2 steps,

Thus:

0 steps -> ①

1 steps -> ②，③，④，⑤

2 steps -> ①，②，③，④，⑥，⑦，⑧

more than 2 steps -> ①, ②, ③, ④, ⑤, ⑥, ⑦, ⑧