Problem:
Score : 400 points

Problem 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.

Constraints

2 \leq n \leq 10^5

1 \leq m \leq 10^5

1 \leq s,t \leq n

s \neq t

1 \leq u_i < v_i \leq n

1 \leq a_i,b_i \leq 10^9

If i\neq j, then u_i \neq u_j or v_i \neq v_j.

Any city can be reached from any city by changing trains.

All values in input are integers.

Input

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

Output

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.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class PriorityQueue
  MIN = -(1 << 60)

  attr_reader :size

  def initialize(&block)
    @size = 0
    @heap = Array.new
    @get_key = block || :itself.to_proc
  end

  def empty?
    size == 0
  end

  def pop
    return nil if @size <= 0
    max = @heap[1]
    @heap[1] = @heap[@size]
    @size -= 1
    max_heapify(1)
    max
  end

  def push(x)
    @size += 1
    i = @size
    @heap[i] = x
    while i > 1 && (@get_key.call(@heap[i / 2]) <=> @get_key.call(@heap[i])) < 0
      t = @heap[i / 2]
      @heap[i / 2] = @heap[i]
      @heap[i] = t
      i = i / 2
    end
  end

  def top
    return nil if @size <= 0
    @heap[1]
  end

  def to_a
    return [] if @size == 0
    @heap[1..@size]
  end

  private

  def max_heapify(i)
    l = 2 * i
    r = 2 * i + 1
    largest =
      l <= @size && (@get_key.call(@heap[l]) <=> @get_key.call(@heap[i])) > 0 ? l : i
    largest = r \
      if r <= @size && (@get_key.call(@heap[r]) <=> @get_key.call(@heap[largest])) > 0
    unless largest == i
      t = @heap[i]
      @heap[i] = @heap[largest]
      @heap[largest] = t
      max_heapify(largest)
    end
  end
end

N, M, S, T = gets.split.map(&:to_i)

yen_graph = Array.new(N) { [] }
snuuk_graph = Array.new(N) { [] }
M.times do
  u, v, a, b = gets.split.map(&:to_i)
  u -= 1
  v -= 1
  yen_graph[u].push([v, a])
  yen_graph[v].push([u, a])
  snuuk_graph[u].push([v, b])
  snuuk_graph[v].push([u, b])
end

def shortest_path(graph, s)
  w_list = Array.new(graph.size, 1 << 60)
  w_list[s] = 0
  queue = PriorityQueue.new do |_, c|
    -c
  end
  queue.push([s, 0])
  while !queue.empty?
    u, w = queue.pop
    next if w > w_list[u]
    graph[u].each do |v, c|
      if w_list[u] + c < w_list[v]
        new_cost = w_list[u] + c
        w_list[v] = new_cost
        queue.push([v, new_cost])
      end
    end
  end
  w_list
end

yen_cost = shortest_path(yen_graph, S - 1)
snuuk_cost = shortest_path(snuuk_graph, T - 1)
cost_list = N.times.map { |i| yen_cost[i] + snuuk_cost[i] }
FIRST_YEN = 10 ** 15
current_cost = 1 << 60
ans = Array.new(N)
(N - 1).downto(0) do |i|
  current_cost = current_cost > cost_list[i] ? cost_list[i] : current_cost
  ans[i] = FIRST_YEN - current_cost
end
ans.each do |a|
  puts a
end