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: # -*- coding: utf-8 -*-
  import sys
  from heapq import heappop, heappush
  from itertools import accumulate
  
  
  def dijkstra(n, edges, start, *, adj_matrix=False, default_value=float("inf")):
  
      inf = float("inf")
      costs = [inf] * n
      costs[start] = 0
      pq, rem = [(0, start)], n - 1
  
      while pq and rem:
          tmp_cost, tmp_node = heappop(pq)
          if costs[tmp_node] < tmp_cost:
              continue
          rem -= 1
  
          nxt_edges = (
              (
                  (node, cost)
                  for node, cost in enumerate(edges[tmp_node])
                  if cost != default_value
              )
              if adj_matrix
              else edges[tmp_node]
          )
  
          for nxt_node, nxt_cost in nxt_edges:
              new_cost = tmp_cost + nxt_cost
              if costs[nxt_node] > new_cost:
                  costs[nxt_node] = new_cost
                  heappush(pq, (new_cost, nxt_node))
  
      return costs
  
  
  # n, m, s, t = map(int, input().split())
  # trains = [list(map(int, l.strip().split())) for l in sys.stdin.readlines()]
  
  # yen_edges = [[float('inf') if __ != _ else 0 for __ in range(n)]
  #              for _ in range(n)]
  # snuke_edges = [[float('inf') if __ != _ else 0 for __ in range(n)]
  #                for _ in range(n)]
  # for i, j, yen, snuke in trains:
  #     yen_edges[i - 1][j - 1] = yen
  #     yen_edges[j - 1][i - 1] = yen
  #     snuke_edges[i - 1][j - 1] = snuke
  #     snuke_edges[j - 1][i - 1] = snuke
  
  # yen_cost = dijkstra(n, yen_edges, s - 1, adj_matrix=True)
  # snuke_cost = dijkstra(n, snuke_edges, t - 1, adj_matrix=True)
  # costs = [yen + snuke for yen, snuke in zip(yen_cost, snuke_cost)]
  
  n, m, s, t = map(int, input().split())
  
  yen_edges = [[] for _ in range(n)]
  snuke_edges = [[] for _ in range(n)]
  for l in sys.stdin.readlines():
      u, v, yen, snuke = map(int, l.strip().split())
      # for _ in range(m):
      #     u, v, yen, snuke = map(int, input().split())
      yen_edges[u - 1].append((v - 1, yen))
      yen_edges[v - 1].append((u - 1, yen))
      snuke_edges[u - 1].append((v - 1, snuke))
      snuke_edges[v - 1].append((u - 1, snuke))
  
  yen_costs = dijkstra(n, yen_edges, s - 1)
  snuke_costs = dijkstra(n, snuke_edges, t - 1)
  
  final_costs = list(
      accumulate([yen + snuke for yen, snuke in zip(yen_costs, snuke_costs)][::-1], min)
  )[::-1]
  asset = 10**15
  for final_cost in final_costs:
      print(asset - final_cost)
  
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"}]