Question
Two Python-style lists arrive as literal assignments. Collect every value that belongs to the first list but not the second, keeping the first list's order. ```python def solve(raw: str) -> str: ... ``` ## Requirements - The input holds two assignment statements, one per entry: `list1 = [...]` followed by `list2 = [...]`. Both lists contain integers and may be empty. - The statements may be separated by a real newline or by a literal two-character `\n` escape; split on either form. - Return the kept values in their original `list1` order. - Membership in `list2` is decided by value, not by count or position, so a single occurrence there removes every matching occurrence in `list1`. - Values that occur only in `list1` are emitted once per occurrence, so duplicates in `list1` survive together. - Print the result exactly as a Python-style list: square brackets, comma plus space, no trailing comma, for example `[1, 3, 5]`. An empty first list answers `[]`. - `run_case` passes `case["stdin"]` to `solve`. - Do not append a trailing newline to the returned string; return the value exactly once.