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: inp = list(map(int, input().split()))
  n = inp[0] # the num of node
  m = inp[1] # the num of edge
  s = inp[2] - 1 # start
  t = inp[3] - 1 # stop
  
  u = [] # edge (smaller)
  v = [] # edge (larger)
  a = [] # fare (yen)
  b = [] # fare (snuke)
  for i in range(m):
      inp = list(map(int, input().split()))
      u.append(inp[0] - 1)
      v.append(inp[1] - 1)
      a.append(inp[2])
      b.append(inp[3])
  
  uu = [[] for i in range(n)]
  
  for i in range(m):
      uu[u[i]].append([v[i], a[i], b[i]])
      uu[v[i]].append([u[i], a[i], b[i]])
  
  # Dijkstra's Algorithm
  #   fare: 1 => yen, 2 => snuke
  def cheaper(start, stop, fare):
  
      if (start == stop):
          return 0
  
      flag = [0 for i in range(n)]
      distance = [-1 for i in range(n)]
      flag[start] = 1
      distance[start] = 0
      base = start
      changeList = set([])
  
      for i in range(n - 1):
          # update distance
          for j in len(uu[base]):
                  befo = distance[uu[base][j][0]]
                  afte = distance[base] + uu[base][j][fare]
                  if (afte < befo or befo == -1):
                      distance[uu[base][j][0]] = afte
                      changeList.add(uu[base][j][0]])
  
          # update base
          base = list(changeList)[0]
          for j in changeList:
              if (distance[j] < distance[base]):
                  base = j
          changeList.remove(base)
  
          # update flag (using updated base)
          flag[base] = 1
  
          # check break
          if (base == stop):
              break
  
      return distance[base]
  
  totalFare = []
  for ex in range(n): # ex: exchange
      totalFare.append(cheaper(s, ex, a) + cheaper(ex, t, b))
  
  for i in range(n):
      money = 10 ** 15
      print(money - min(totalFare))
      del totalFare[0]
  
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"}]