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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import scala.collection.mutable.{ListBuffer, PriorityQueue}

object Main extends App {

  val init = 1e15.toLong

  val in = io.StdIn.readLine().split(' ')
  val (n, m, s, t) = (in(0).toInt, in(1).toInt, in(2).toInt - 1, in(3).toInt - 1)

  val g = Seq.fill(n)(ListBuffer[(Int, Long, Long)]())

  (1 to m).map(_ => io.StdIn.readLine().split(' '))
    .map(in => (in(0).toInt - 1, in(1).toInt - 1, in(2).toLong, in(3).toLong))
    .foreach { case (u, t, a, b) =>
      g(u) += ((t, a, b))
      g(t) += ((u, a, b))
    }

  val yenSeq = dijkstra(s, g, { case (_, a, _) => a })
  val snkSeq = dijkstra(t, g, { case (_, _, b) => b })
  val ans = Array.fill[Long](n + 1)(0L)

  Range(1, n + 1)
    .map(n - _)
    .foreach(idx => ans(idx) = math.max(ans(idx + 1), init - yenSeq(idx) - snkSeq(idx)))

  Range(0, n).foreach(idx => println(ans(idx)))

  def dijkstra(s: Int, g: Seq[Seq[(Int, Long, Long)]], f: ((Int, Long, Long)) => Long): Array[Long] = {
    val pq = new PriorityQueue[(Long, Int)]()(Ordering.by[(Long, Int), Long](-_._1))

    val score = Array.fill(n)(1L << 50)
    score(s) = 0
    pq.enqueue((0L, s))

    while (pq.nonEmpty) {
      val (c, p) = pq.dequeue
      for {
        tup <- g(p)
        if score(tup._1) > c + f(tup)
      } yield {
        score(tup._1) = c + f(tup)
        pq.enqueue((c + f(tup), tup._1))
      }
    }
    score
  }
}