output_description: If the graph contains a negative cycle (a cycle whose sum of edge costs is a
  negative value) which is reachable from the source r, print
  
      
      
      NEGATIVE CYCLE
      
  
  in a line.
  
  Otherwise, print
  
      
      
      c0
      c1
      :
      c|V|-1
      
  
  The output consists of |V| lines. Print the cost of the shortest path from the
  source r to each vertex 0, 1, ... |V|-1 in order. If there is no path from the
  source to a vertex, print "INF".
input_description: An edge-weighted graph G (V, E) and the source r.
  
      
      
      |V| |E| r
      s0 t0 d0
      s1 t1 d1
      :
      s|E|-1 t|E|-1 d|E|-1
      
  
  |V| is the number of vertices and |E| is the number of edges in G. The graph
  vertices are named with the numbers 0, 1,..., |V|-1 respectively. r is the
  source of the graph.
  
  si and ti represent source and target vertices of i-th edge (directed) and di
  represents the cost of the i-th edge.
attempt: """
  
  
  Verified:
      https://onlinejudge.u-aizu.ac.jp/problems/GRL_1_B
  """
  
  from typing import List, Optional, Tuple
  
  
  def bellman_ford(
      source: int, vertices_count: int, edges: List[Tuple]
  ) -> Optional[List[float]]:
      """Bellman-Ford algorithm: O(NM)
      Compute the distance of the shortest path from V_source to each vertex.
      Relax edges in N-1 times of loops and detect negative cycles in the last loop.
  
      Args:
          source: V_source
          edges: Edges info
          vertices_count: Number of vertices
      Returns:
          Return None if the graph has negative cycles.
          If not, return a list of distance from V_source.
      """
      inf = float("inf")
      dist = [inf] * (vertices_count + 1)
      dist[source] = 0
      for i in range(vertices_count):
          for from_, to, weight in edges:
              if dist[from_] + weight < dist[to]:
                  dist[to] = dist[from_] + weight
                  if i == vertices_count - 1:
                      return None
      return dist
  
  
  def main():
      V, E, R, *edges = map(int, open(0).read().split())
      dist = bellman_ford(R, V, list(zip(*[iter(edges)] * 3)))
      inf = float("inf")
      if dist:
          print("\n".join(map(str, (i if i != inf else "INF" for i in dist[:-1]))))
      else:
          print("NEGATIVE CYCLE")
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Single Source Shortest Path (Negative Edges)
samples: [{"input": "4 5 0\n    0 1 2\n    0 2 3\n    1 2 -5\n    1 3 1\n    2 3 2", "output": "0\n    2\n    -3\n    -1"}, {"input": "4 6 0\n    0 1 2\n    0 2 3\n    1 2 -5\n    1 3 1\n    2 3 2\n    3 1 0", "output": "NEGATIVE CYCLE"}, {"input": "4 5 1\n    0 1 2\n    0 2 3\n    1 2 -5\n    1 3 1\n    2 3 2", "output": "INF\n    0\n    -5\n    -3"}]