Why does this algorithm work in O(n m)?

This is from a blog post on Codeforces. I couldn't really understand why the editorialist goes on to claim that this code works in O(n m)

This is a graph problem, where we are supposed to find the number of ways to traverse from a to c. Note that only the paths like a-->b-->c and a-->d-->c need to be considered. That is, there should be a difference of only one node, in between them.

Let's iterate through all combinations of a and c just two simple nested loops in O(n^2) and find all candidates for b and d inside. To find candidates you can go through all neighbors of a and check that they are neighbors of c. Among all the candidates you should choose two junctions as b and d. So just use http://ift.tt/1vZTgjE All you need is to add to the answer , where r is the number of candidates (common neighbors of a and c). The code is:

for (int a = 0; a < n; a++)
    for (int c = 0; c < n; c++)
        if (a != c)
        {
            int r = 0;
            for (int b = 0; b < nxt[a].size(); b++)
                if (nxt[a][b] != a && nxt[a][b] != c && g[nxt[a][b]][c])
                    r++;
            result += r * (r - 1) / 2;
        }

It is easy to see that the total complexity is O(nm), because of sum of number of neighbors over all junctions is exactly m.

If there are 3 loops involved, then how can the worst case complexity be O(n m)