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 java.io.{BufferedReader, InputStream, InputStreamReader, PrintWriter}
import java.util.StringTokenizer

import scala.collection.mutable
import scala.util.Sorting
import math.{abs, max, min}
import scala.collection.mutable.{ArrayBuffer, ListBuffer}
import scala.reflect.ClassTag

object Main {
  val MOD = 1000000007
  val out = new PrintWriter(System.out)
  def solve(): Unit = {
    val sc = new InputReader(System.in)
    val n, m = sc.nextInt()
    val s, t = sc.nextInt() - 1
    type Node = Int
    type Edge = (Node ,Long)
    val t1, t2 = map(n)(_ => ListBuffer.empty[Edge])
    rep(m) { _ =>
      val u, v = sc.nextInt() - 1
      val a, b = sc.nextLong()
      t1(u) += ((v, a))
      t1(v) += ((u, a))
      t2(u) += ((v, b))
      t2(v) += ((u, b))
    }

    def dijkstra(tree: Array[ListBuffer[Edge]], start: Int): Array[Long] = {
      val d = Array.fill[Long](n)(Long.MaxValue)
      type Visit = (Node, Long)
      val queue = mutable.PriorityQueue.empty[Visit]
      d(start) = 0
      tree(start) foreach { case (u, c) => queue.enqueue((u, c)) }

      while(queue.nonEmpty) {
        val (v, c0) = queue.dequeue()
        if (d(v) > c0) {
          d(v) = c0
          tree(v) foreach { case (u, c1) =>
            if (d(u) > c0 + c1) queue.enqueue((u, c0 + c1))
          }
        }
      }
      
      d
    }

    val d1 = dijkstra(t1, s)
    val d2 = dijkstra(t2, t)

    val d = map(n){ i =>
      val j = n - 1 - i
      d1(j) + d2(j)
    }
    rep(n - 1) { i =>
      d(i + 1) = min(d(i + 1), d(i))
    }
    val ans = map(n){ i => 1000000000000000L - d(i) }
    ans.reverse foreach out.println
  }


  def main(args: Array[String]): Unit = {
    solve()
    out.flush()
  }


  class InputReader(val stream: InputStream) {
    private val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private var tokenizer: StringTokenizer = null

    def next: String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = next.toInt
    def nextLong(): Long = next.toLong
    def nextChar(): Char = next.charAt(0)
  }

  private def rep(n: Int)(f: Int => Unit): Unit = (0 until n) foreach f

  private def map[@specialized A: ClassTag](n: Int)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    (0 until n) foreach (i => res(i) = f(i))
    res
  }

  private implicit class ArrayOpts[A](val as: Array[A]) extends AnyVal {
    def maxByOpt[B: Ordering](f: A => B): Option[A] = {
      if (as.nonEmpty) Some(as.maxBy(f)) else None
    }

    def grpBy[K](f: A => K): mutable.Map[K, ArrayBuffer[A]] = {
      val map = mutable.Map.empty[K, ArrayBuffer[A]]
      rep(a => map.getOrElseUpdate(f(a), ArrayBuffer()) += a)
      map
    }

    def rep(f: A => Unit): Unit = {
      as.indices foreach (i => f(as(i)))
    }

    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      var sum = num.zero
      rep(a => sum = num.plus(sum, f(a)))
      sum
    }

    def minByEx[B](f: A => B, ixRange: Range = as.indices)(implicit cmp: Ordering[B]): (A, B) = {
      limit(f, ixRange)(cmp.lt)
    }

    def maxByEx[B](f: A => B, ixRange: Range = as.indices)(implicit cmp: Ordering[B]): (A, B) = {
      limit(f, ixRange)(cmp.gt)
    }

    private def limit[B](f: A => B, ixRange: Range)(cmp: (B, B) => Boolean): (A, B) = {
      var limA = as(ixRange.head)
      var limB = f(limA)

      for (i <- ixRange.tail) {
        val a = as(i)
        val b = f(a)
        if (cmp(b, limB)) {
          limA = a
          limB = b
        }
      }
      (limA, limB)
    }
  }

  private implicit class IterableOpts[A](val as: Iterable[A]) extends AnyVal {
    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      as.foldLeft(num.zero)((acc, a) => num.plus(acc, f(a)))
    }
  }
}