How can I find and compute the K-shortest disjoint pair paths between two nodes in a graph?

Question

I want to find and compute the K-shortest pair paths between two nodes in which each pair path contains two paths that they have not any shared links( only source and target nodes are common in each pair so they are disjoint and considered as a disjoint pair path). now I wanna compute the K number of those disjoint pair paths between two nodes.
my idea is like this: first of all, I compute K-shortest paths between two nodes, and after that, I remove the shortest one between them from the graph and I apply K-shortest paths between the same two nodes again. now I have the first removed shortest path and some other paths that I reached after the second applying of k-shortest paths between the same nodes and they are disjoint.
in the other words actually, I have one path with K other paths that have not any shared link with the first(removed) path. I want to repeat this procedure for many requests in the graph(any given two nodes in the graph). I wrote my own code but that didn't work properly for me! because the paths in output are not disjoint.
here is a part of my code:

for each request r:  
  def k_shortest_paths(G, source, target, k, weight='weight'):
    return list(islice(nx.shortest_simple_paths(G, source, target, weight=weight), k))

  i=1  
  for path in k_shortest_paths(G, source, target, 2):
     print("working",i,":",path)
     working=k_shortest_paths(G, source, target, 1)
     sp=[]
     for i in range(0,len(working)-1):
       sp.append((working[i],working[i+1])) 
     for u, v in sp:
       G.remove_edge(u, v)
   
     for path in k_shortest_paths(G, source, target, 2):
       print("protection",i,":",path)
  
       i=i+1

with this code, I reach two paths as 'working' paths and two other paths(for each working path) as 'protection' paths which working and protection paths are disjoint. my goal is to find K-protection paths for each working path. I defined my graph using networkx library. I hope the explanation is sufficient!
Does anybody have any solution or maybe a simpler way to my problem? thanks in advance.