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 Updatable_Binary_Heap
  def initialize(list: [], &pred)
    @a = [nil] + list.sort{ |a,b| pred.call(a,b) ? -1 : 1 }
    @pred = pred
    @m = @a.size
    @position = {}
    @a[1..-1].each_with_index{ |v,i| @position[v] = i+1 }
  end

  def up_heap(i)
    while (j = i >> 1) > 0
      break if @pred.call(@a[j], @a[i])
      swap_node(i,j)
      i = j
    end
    i
  end

  def head
    @a[@m > 1 ? 1 : 0]
  end

  def shift
    return nil if @m <= 1
    @m -= 1
    swap_node(1, @m)
    down_heap(1)
    @a[@m]
  end

  def down_heap(i)
    while (j = i << 1) < @m
      j |= ((j|1) < @m && !@pred.call(@a[j], @a[j|1])) ? 1 : 0
      break if @pred.call(@a[i], @a[j])
      swap_node(i,j)
      i = j
    end
    i
  end

  def swap_node(i,j)
    @a[i], @a[j] = @a[j], @a[i]
    @position[@a[i]] = i
    @position[@a[j]] = j
  end

  def update_node(node)
    i = @position[node]
    down_heap(i) == i && up_heap(i)
  end

  def a
    @a[1...@m]
  end
end

def shortest_dijkstra(nodes, edges, start, infinity: 1 << 32)
  costs = nodes.map{ |n| n == start ? 0 : infinity }
  ns = Updatable_Binary_Heap.new(list: nodes) { |s,t| costs[s-1] <= costs[t-1] }
  vs = ns.a.sort_by{ |v| v }
  #p ns
  #p vs
  determined = {}
  cnt = 0
  while ui = ns.shift
    nu = ui
    u = costs[nu-1]
    cnt += 1
    break if u >= infinity
    next if determined[nu]
    determined[nu] = 1
    next unless h = edges[nu]
    h.each{ |nv,w|
      vi = vs[nv-1]
      v = costs[nv-1]
      next if v <= u + w
      v = u + w
      costs[nv-1] = v
      #ns.insert(v)
      ns.update_node(vi)
      #p ns.head
    }
  end
  #$stderr.puts cnt, nodes.size
  costs
end

n,m,s,t = gets.split.map(&:to_i)

rails_yn = {}
rails_sn = {}
(0...m).each{ 
  v,u,a,b = gets.split.map(&:to_i) 
  rails_sn[v] ||= {}; rails_sn[u] ||= {}
  rails_yn[v] ||= {}; rails_yn[u] ||= {}
  rails_yn[v][u] = rails_yn[u][v] = a
  rails_sn[v][u] = rails_sn[u][v] = b
}
ns = (1..n).to_a
INF = 1000000000000000

yh = shortest_dijkstra(ns, rails_yn, s, infinity: INF)
sh = shortest_dijkstra(ns, rails_sn, t, infinity: INF)
#p yh, sh
costs = (0...n).map{ |i| yh[i] + sh[i] }

heap = Updatable_Binary_Heap.new(list: (1..n).to_a) { |s,t| costs[s-1] <= costs[t-1] }
heap_nodes = heap.a.sort_by{ |n| n }

#p costs
#p heap
(1..n).each{ |i|
  h = heap.head
  puts INF - costs[h-1]
  u = heap_nodes[i-1]
  #p u
  costs[i-1] = INF
  heap.update_node(u)
}