output_description: Print n lines. In the i-th line, print the maximum amount of snuuk that
  Kenkoooo has when he reaches City t if he goes on a trip from City s to City t
  after i-1 years.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      n m s t
      u_1 v_1 a_1 b_1
      :
      u_m v_m a_m b_m
attempt: import sys
  import math
  import collections
  import itertools
  import array
  import inspect
  
  # Set max recursion limit
  sys.setrecursionlimit(10000)
  
  
  # Debug output
  def chkprint(*args):
      names = {id(v): k for k, v in inspect.currentframe().f_back.f_locals.items()}
      print(", ".join(names.get(id(arg), "???") + " = " + repr(arg) for arg in args))
  
  
  # Binary converter
  def to_bin(x):
      return bin(x)[2:]
  
  
  # --------------------------------------------
  
  dp = None
  
  
  def main():
      n_station, n_train, src_station, dst_station = [int(_) for _ in input().split()]
  
      connect = []
      for _ in range(n_station + 1):
          connect.append([])
  
      trains = {}
      for _ in range(n_train):
          u, v, a, b = [int(_) for _ in input().split()]
          trains[str(u) + str(v)] = {"yen": a, "snk": b}
          connect[u].append(v)
          connect[v].append(u)
  
      costs_by_exchange_station = [10**20]
  
      stat_and_cost_arr = [(src_station, 0)]
      costs_yen = {src_station: 0}
      while True:
          new_stat_and_cost_arr = []
          for stat_and_cost in stat_and_cost_arr:
              new_stat_and_cost_arr = []
              current_station, current_cost = stat_and_cost[0], stat_and_cost[1]
  
              for next_station in connect[current_station]:
                  u, v = min(current_station, next_station), max(
                      current_station, next_station
                  )
                  if (
                      next_station not in costs_yen
                      or current_cost + trains[str(u) + str(v)]["yen"]
                      < costs_yen[next_station]
                  ):
                      new_stat_and_cost_arr.append(
                          (next_station, current_cost + trains[str(u) + str(v)]["yen"])
                      )
                      costs_yen[next_station] = (
                          current_cost + trains[str(u) + str(v)]["yen"]
                      )
  
          if len(new_stat_and_cost_arr) == 0:
              break
          else:
              stat_and_cost_arr = new_stat_and_cost_arr
  
      for exchange_station in range(1, n_station + 1):
          stat_and_cost_arr = [(exchange_station, costs_yen[exchange_station])]
  
          costs_snk = {exchange_station: costs_yen[exchange_station]}
          while True:
              new_stat_and_cost_arr = []
              for stat_and_cost in stat_and_cost_arr:
                  new_stat_and_cost_arr = []
                  current_station, current_cost = stat_and_cost[0], stat_and_cost[1]
  
                  for next_station in connect[current_station]:
                      u, v = min(current_station, next_station), max(
                          current_station, next_station
                      )
                      if (
                          next_station not in costs_snk
                          or current_cost + trains[str(u) + str(v)]["snk"]
                          < costs_snk[next_station]
                      ):
                          new_stat_and_cost_arr.append(
                              (
                                  next_station,
                                  current_cost + trains[str(u) + str(v)]["snk"],
                              )
                          )
                          costs_snk[next_station] = (
                              current_cost + trains[str(u) + str(v)]["snk"]
                          )
  
              if len(new_stat_and_cost_arr) == 0:
                  break
              else:
                  stat_and_cost_arr = new_stat_and_cost_arr
  
          costs_by_exchange_station.append(costs_snk[dst_station])
  
      for year in range(n_station):
          print(10**15 - min(costs_by_exchange_station[year + 1 :]))
  
  
  main()
  
problem_description: Statement
  
  Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n
  cities and m trains running. The cities are numbered 1 through n, and the i-th
  train connects City u_i and v_i bidirectionally. Any city can be reached from
  any city by changing trains.
  
  Two currencies are used in the country: yen and snuuk. Any train fare can be
  paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in
  yen, and b_i snuuk if paid in snuuk.
  
  In a city with a money exchange office, you can change 1 yen into 1 snuuk.
  However, when you do a money exchange, you have to change all your yen into
  snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will
  then have X snuuk. Currently, there is a money exchange office in every city,
  but the office in City i will shut down in i years and can never be used in
  and after that year.
  
  Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head
  for City t, and change his yen into snuuk in some city while traveling. It is
  acceptable to do the exchange in City s or City t.
  
  Kenkoooo would like to have as much snuuk as possible when he reaches City t
  by making the optimal choices for the route to travel and the city to do the
  exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo
  has when he reaches City t if he goes on a trip from City s to City t after i
  years. You can assume that the trip finishes within the year.
samples: [{"input": "4 3 2 3\n    1 4 1 100\n    1 2 1 10\n    1 3 20 1", "output": "999999999999998\n    999999999999989\n    999999999999979\n    999999999999897\n    \n\nAfter 0 years, it is optimal to do the exchange in City 1.  \nAfter 1 years, it is optimal to do the exchange in City 2.  \nNote that City 1 can still be visited even after the exchange office is\nclosed. Also note that, if it was allowed to keep 1 yen when do the exchange\nin City 2 and change the remaining yen into snuuk, we could reach City 3 with\n999999999999998 snuuk, but this is NOT allowed.  \nAfter 2 years, it is optimal to do the exchange in City 3.  \nAfter 3 years, it is optimal to do the exchange in City 4. Note that the same\ntrain can be used multiple times.\n\n* * *"}, {"input": "8 12 3 8\n    2 8 685087149 857180777\n    6 7 298270585 209942236\n    2 4 346080035 234079976\n    2 5 131857300 22507157\n    4 8 30723332 173476334\n    2 6 480845267 448565596\n    1 4 181424400 548830121\n    4 5 57429995 195056405\n    7 8 160277628 479932440\n    1 6 475692952 203530153\n    3 5 336869679 160714712\n    2 7 389775999 199123879", "output": "999999574976994\n    999999574976994\n    999999574976994\n    999999574976994\n    999999574976994\n    999999574976994\n    999999574976994\n    999999574976994"}]