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: # -*- coding: utf-8 -*-
  
  import sys
  
  sys.setrecursionlimit(10**9)
  
  
  def input():
      return sys.stdin.readline().strip()
  
  
  def INT():
      return int(input())
  
  
  def MAP():
      return map(int, input().split())
  
  
  def LIST():
      return list(map(int, input().split()))
  
  
  INF = float("inf")
  
  
  # ベルマンフォード(頂点数, 辺集合(0-indexed), 始点)
  def bellman_ford(N: int, edges: list, src: int) -> list:
      # 頂点[ある始点からの最短距離]
      # (経路自体を知りたい時はここに前の頂点も持たせる)
      res = [INF] * N
      res[src] = 0
      # 各辺によるコストの置き換えを頂点数N-1回繰り返す
      for i in range(N - 1):
          for src, dest, cost in edges:
              if res[dest] > res[src] + cost:
                  res[dest] = res[src] + cost
      # 負の閉路(いくらでもコストを減らせてしまう場所)がないかチェックする
      for src, dest, cost in edges:
          if res[dest] > res[src] + cost:
              # あったら空リストを返却
              return []
      # 問題なければ頂点リストを返却
      return res
  
  
  N, M, r = MAP()
  edges = [None] * M
  for i in range(M):
      s, t, d = MAP()
      edges[i] = (s, t, d)
  
  ans = bellman_ford(N, edges, r)
  
  if not len(ans):
      print("NEGATIVE CYCLE")
      exit()
  
  for i in range(N):
      if ans[i] == INF:
          ans[i] = "INF"
  for i in range(N):
      print(ans[i])
  
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"}]