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)

NODE = Struct.new(:v, :value) do
  def <=>(node)
    value <=> node.value
  end
end

class FastPriorityQueue
  attr_reader :size

  def initialize
    @size = 0
    @nodes = []
  end

  def empty?
    @size.zero?
  end

  def push(node)
    index = size
    @size += 1

    while index > 0
      parent = (index - 1) / 2

      break if (@nodes[parent] <=> node) < 0

      @nodes[index] = @nodes[parent]
      index = parent
    end

    @nodes[index] = node
  end

  alias_method :<<, :push

  def pop
    node = @nodes.first
    root = @nodes[size - 1]
    @nodes[size - 1] = nil
    @size -= 1

    return node if empty?

    index = 0

    while index * 2 + 1 < @size
      left = 2 * index + 1
      right = left + 1

      if right < size && (@nodes[right] <=> @nodes[left]) < 0
        left = right
      end

      break if (@nodes[left] <=> root) >= 0

      @nodes[index] = @nodes[left]
      index = left
    end

    @nodes[index] = root
    node
  end

  def top
    @nodes.first
  end
end

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

E_EN = Hash.new { |h, k| h[k] = Hash.new(Float::INFINITY) }
E_SNUKU = Hash.new { |h, k| h[k] = Hash.new(Float::INFINITY) }

M.times do
  u, v, a, b = gets.split.map(&:to_i)

  E_EN[u][v] = E_EN[v][u] = a
  E_SNUKU[u][v] = E_SNUKU[v][u] = b
end

pque = FastPriorityQueue.new

pque.push(NODE.new(S, 0))
before_cost = Hash.new

until pque.empty?
  node = pque.pop

  next if before_cost[node.v]
  before_cost[node.v] = node.value

  E_EN[node.v].each do |u, cost|
    pque.push(NODE.new(u, node.value + cost))
  end
end

pque.push(NODE.new(T, 0))
after_cost = Hash.new

until pque.empty?
  node = pque.pop

  next if after_cost[node.v]
  after_cost[node.v] = node.value

  E_SNUKU[node.v].each do |u, cost|
    pque.push(NODE.new(u, node.value + cost))
  end
end

costs = Hash.new

(1..N).each do |m|
  cost = before_cost[m] + after_cost[m]

  costs[m] = cost
end

list = costs.sort_by { |m, c| c }
money = 10 ** 15

N.times do |y|
  while list[0][0] <= y
    list.shift
  end

  puts money - list[0][1]
end