**Firstly, we may take n = min(n, 3).** The sequence of lights clearly repeats every 6, so the first 6 lights are representative of the whole sequence, as we can construct eg. the 7th light (it's equal to the 1st).

Actually, the first 3 lights are representative of the whole sequence. If the operations are a, b, c, d; then modulo 2:

- Light 1 = 1 + a + c + d
- Light 2 = 1 + a + b
- Light 3 = 1 + a + c
- Light 4 = 1 + a + b + d
- Light 5 = 1 + a + c
- Light 6 = 1 + a + b

So that (modulo 2):

- Light 4 = (Light 1) + (Light 2) + (Light 3)
- Light 5 = Light 3, and
- Light 6 = Light 2.

Now, we can do cases on `m`

, and analyze the possible lightbulb states for `n >= 3`

. The transitions are to XOR by (1, 1, 1), (0, 1, 0), (1, 0, 1), or (1, 0, 0).

- If
`m = 0`

there is only one state`(1, 1, 1)`

. - If
`m = 1`

then we could get`(0, 0, 0), (1, 0, 1), (0, 1, 0), (0, 1, 1)`

. - If
`m = 2`

we could get all 8 possibilities except`(0, 1, 1)`

. - If
`m = 3`

we can get every possibility.

This reduced the problem to knowing the answer for `m <= 3, n <= 3`

. The final answer is:

- When
`n == 1`

, the answer is 1 if`m == 0`

, else 2. - When
`n == 2`

, the answer is 1 if`m == 0`

, 3 if`m == 1`

, else 4. - When
`n >= 3`

, the answer is 1 if`m == 0`

, 4 if`m == 1`

, 7 if`m == 2`

, else 8.