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
  include Enumerable

  MIN = -(2 << 60)

  attr_reader :size

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

  def each
    (1..@size).each do |i|
      yield @heap[i]
    end
  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])
      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

  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]) ? l : i
    largest = r \
      if r <= @size && @get_key.call(@heap[r]) > @get_key.call(@heap[largest])
    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)
uvab_list = M.times.map { gets.split.map(&:to_i) }
MONEY = 10 ** 15
yen_graph = Array.new(N) { Array.new }
snuke_graph = Array.new(N) { Array.new }
uvab_list.each do |(u, v, a, b)|
  u -= 1
  v -= 1
  yen_graph[u].push([v, a])
  yen_graph[v].push([u, a])
  snuke_graph[u].push([v, b])
  snuke_graph[v].push([u, b])
end

def dijkstra(s, graph)
  pq = PriorityQueue.new { |(_, d)| -d }
  d_list = Array.new(graph.size, 1 << 60)
  d_list[s] = 0
  pq.push([s, 0])
  while !pq.empty?
    v, d = pq.pop
    next if d > d_list[v]
    graph[v].each do |(u, c)|
      if d_list[u] > d_list[v] + c
        d_list[u] = d_list[v] + c
        pq.push([u, d_list[u]])
      end
    end
  end
  d_list
end

yen_d_list = dijkstra(S - 1, yen_graph)
snuke_d_list = dijkstra(T - 1, snuke_graph)
cost_list = yen_d_list.zip(snuke_d_list).map { |(y, s)| y + s }
ans_list = Array.new(N)
ans_list[N - 1] = MONEY - cost_list[N - 1]
(N - 2).downto(0).each do |i|
  ans_list[i] = [MONEY - cost_list[i], ans_list[i + 1]].max
end
ans_list.each { |ans| puts ans }