在这种情况下如何摆脱未分配的指针？

Question

以下是我使用的事实：

edge(route(fra), route(tog), walk).
edge(route(fra), route(105), walk).
edge(route(tog), route(togc), bus).
edge(route(togc), route(sub), walk).
edge(route(sub), route(ag), metro).
edge(route(ag), route(n), metro).
edge(route(n), route(a), metro).
edge(route(a), route(r), metro).
edge(route(re), route(m), metro).
edge(route(m), route(v), walk).
edge(route(105), route(t), bus).
edge(route(t), route(vi), metro).
edge(route(vi), route(m), metro).
edge(route(t), route(3), walk).
edge(route(3), route(m), bus).

谓词：

seek(route(v), _).
seek(route(R1), [[R2, How]|Res]) :-
    edge(route(R1), route(R2), How),
    seek(route(R2),Res).

当我询问查询?- seek(route(fra), X).

时

它应该在列表中返回R2和How，但在这种情况下Res仍然未分配...我不知道如何绕过它来进行查询仅返回包含R2的列表列表。任何帮助将不胜感激！