Why the worst runtime is O(|V|^2 + |E|)

Consider this implementation of Dijkstra's algorithm using a SpecialPQ with the following asymptotic runtimes for each operation in terms of N, the size of the priority queue. add is in (N). removeMin is in O(N). contains is in (1). changePriority is in O(1). Which of the following best describes the worst-case runtime bound for this implementation of Dijkstra's algorithm in terms of the number of vertices |V| and the number of edges |E|? dijkstra (Graph graph, Vertex start) { Map distTo = new HashMap(); Map edge To = new HashMap(); PriorityQueue perimeter = new SpecialPQ>(); perimeter.add(start); while (!perimeter.isEmpty()) { Vertex from = perimeter.removeMin(); for (Edge edge : graph.neighbors (from)) { Vertex to = edge.to(); double oldDist = distTo.getOrDefault(to, Double. INFINITY); double newDist = distTo.get(from) + edge.weight(); if (newDist