Finding the path of a negative weight cycle using Bellman-Ford

I wrote a program which implements Bellman-Ford, and identifies when negative weight cycles are present in a graph. However what I'm actually interested in, is given some starting vertex and a graph, which path do I actually trace to get to the original vertex having traveled a negative amount.

So to be clear say I have a graph with vertexes, a, b, c, and d and there is a negative cycle between a, b, and d, then when I check for negative weight cycles

// Step 1: initialize graph
   for each vertex v in vertices:
       if v is source then distance[v] := 0
       else distance[v] := infinity
       predecessor[v] := null

   // Step 2: relax edges repeatedly
   for i from 1 to size(vertices)-1:
       for each edge (u, v) with weight w in edges:
           if distance[u] + w < distance[v]:
               distance[v] := distance[u] + w
               predecessor[v] := u

   // Step 3: check for negative-weight cycles
   for each edge (u, v) with weight w in edges:
       if distance[u] + w < distance[v]:
           "Graph contains a negative-weight cycle"

Instead of it just telling me that a negative cycle is there, I would like it to tell me, go from a -> b -> d -> a. After the relaxing step what do I have to change in my check for negative weight cycles to get it to output this information?

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Here is the best information I've been able to find, but I'm still having trouble making sense of it.

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Also this which suggests that I need to run breadth first search on the predecessor array to find the information, but I'm not exactly sure where to start (what do I queue first?)

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Here is a stack overflow question which shows how to find one of the nodes in the path.

Per Kleinberg–Tardos, you want to run Bellman–Ford for n iterations and find a cycle in the predecessor array.

To find a cycle in the predecessor array, start by coloring every node white. For each node u in an arbitrary order, set v := u, and, while v is white and has a predecessor, recolor v gray and set v := predecessor[v]. Upon exiting the loop, if v is gray, we found a cycle; loop through again to read it off. Otherwise, none of the gray nodes are involved in a cycle; loop through again to recolor them black.

StackExchange Computer Science Q#12129, answer score: 4