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: Accepted)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}
import scala.collection.mutable.PriorityQueue

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val n, m, s, t = in.next().toInt

  val graphA = Array.fill[List[(Int, Int)]](n+1)(Nil)
  val graphB = Array.fill[List[(Int, Int)]](n+1)(Nil)

  for(_ <- 1 to m) {
    val u, v, a, b = in.next().toInt
    graphA(u) = (v, a) :: graphA(u)
    graphA(v) = (u, a) :: graphA(v)
    graphB(u) = (v, b) :: graphB(u)
    graphB(v) = (u, b) :: graphB(v)
  }


  // fro から to への最短距離(v は頂点数)
  def dijkstra(from: Int, to: Int, v: Int, edge: Array[List[(Int, Int)]]): Array[Long] = {
    val MAX = Long.MaxValue

    val dist = Array.fill(v+1)(MAX)   // fro からの距離を格納する
    val check = new Array[Boolean](v+1)  // すでにその頂点からを調べたかのフラグ

    dist(from) = 0

    val pq = new PriorityQueue[(Long, Int)]  // (-cost, node) queue
    pq.enqueue((-0, from))

    // 頂点は v 個あるので、v 周する
    for(_ <- 0 until v) {
      // 調べ済みでない頂点のうち、最も近い頂点を now に入れる
      var nowNode = -1       // このターンにフォーカスするノード

      breakable {
        while(true) {
          nowNode = pq.dequeue._2  // 最小コストから貪欲に取っていく
          if(!check(nowNode)){  // まだチェックされていないノードだったら決定
            check(nowNode) = true
            break
          }
        }
      }

      // その頂点からたどり着ける頂点の情報を更新する
      for((nextNode, cost) <- edge(nowNode)) {
        val nextDist = dist(nowNode) + cost
        if(nextDist < dist(nextNode)){  // 既存の通路より近いなら更新
          dist(nextNode) = nextDist
          pq.enqueue((-nextDist, nextNode))
        }
      }
    }

    dist
  }

  val init = 1e15.toLong

  val dpA = dijkstra(s, t, n, graphA)
  val dpB = dijkstra(t, s, n, graphB)

  val ans = new Array[Long](n+1)
  ans(n) = 0L

  for(i <- (0 until n).reverse) {
    val now = init - (dpA(i+1) + dpB(i+1))
    ans(i) = now max ans(i+1)
  }

  for(i <- 0 until n) {
    pw.println(ans(i))
  }
  pw.flush()
}


class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}