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
  # fin = sys.stdin.readline
  from collections import defaultdict
  from heapq import heappop, heappush
  
  
  def initialize(adj, s):
      MAX_NUM = float('inf')
      parent = {s: None}
      d = {vertex: MAX_NUM for vertex in adj.keys()}
      d[s] = 0
      return d, parent
  
  
  # Dijkstra can work using a minheap which does not have decrease_key.
  # and also we do not have to implement Vertex class.
  # Instead, we can use (distance, vertex).
  # in this case, the time it takes is O(ElgE).
  # E is at most O(V^2), thus O(ElgE) = O(ElgV).
  def dijkstra_simple(adj, w, s):
      d, parent = initialize(adj, s)
      priority_queue = []
      heappush(priority_queue, (d[s], s))
      while priority_queue:
          distance, source = heappop(priority_queue) 
          for neighbor in adj[source]:
              new_distance = distance + w[(source, neighbor)]
              if new_distance < d[neighbor]:
                  d[neighbor] = new_distance
                  parent[neighbor] = source
                  heappush(priority_queue, (new_distance, neighbor))
      return d
  
  
  n, m, s, t = map(int, input().split())
  adj = defaultdict(list)
  yen_weights = dict()
  snuuk_weights = dict()
  for _ in range(m):
      u, v, a, b = map(int, input().split())
      for v1, v2 in [(u, v), (v, u)]:
          adj[v1].append(v2)
          yen_weights[(v1, v2)] = a
          snuuk_weights[(v1, v2)] = b
  
  d_yen = dijkstra_simple(adj, yen_weights, s)
  d_snuuk = dijkstra_simple(adj, snuuk_weights, t)
  min_cost = float('inf')
  min_cost_after_each_year_reversed = []
  for vertex in range(n, 0, -1):
      min_cost = min(min_cost, d_yen[vertex] + d_snuuk[vertex])
      min_cost_after_each_year_reversed.append(min_cost)
  
  initial_money = 10**15
  print(*(initial_money - cost for cost in reversed(min_cost_after_each_year_reversed)), sep='\n')
  in reversed(min_cost_after_each_year_reversed)), sep='\n')
  
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"}]