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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val (n, m, start, goal) = readLine()!!.split(' ').map(String::toInt)
    val yenNodes = mutableMapOf<Int, Node>()
    val snuukNodes = mutableMapOf<Int, Node>()

    for (i in 1..n) {
        yenNodes[i] = Node(i)
        snuukNodes[i] = Node(i)
    }
    for (i in 1..m) {
        val (from, to, yenCost, snuukCost) = readLine()!!.split(' ').map(String::toInt)
        yenNodes[from]!!.edgesTo.add(to)
        yenNodes[from]!!.edgesCost.add(yenCost.toLong())
        yenNodes[to]!!.edgesTo.add(from)
        yenNodes[to]!!.edgesCost.add(yenCost.toLong())

        snuukNodes[from]!!.edgesTo.add(to)
        snuukNodes[from]!!.edgesCost.add(snuukCost.toLong())
        snuukNodes[to]!!.edgesTo.add(from)
        snuukNodes[to]!!.edgesCost.add(snuukCost.toLong())
    }

    calcDijkstraCost(yenNodes, start)
    calcDijkstraCost(snuukNodes, goal)

    for (i in 0 until n) {
        var minCost = Long.MAX_VALUE
        for (j in (i + 1)..n) {
            minCost = Math.min(minCost, yenNodes[j]!!.cost + snuukNodes[j]!!.cost)
        }
        println(1000000000000000 - minCost)
    }
}


class Node(val id:Int) : Comparable<Node> {
    override fun compareTo(other: Node): Int {
        return if (this.id == other.id) 0 else 1
    }

    val edgesTo = mutableListOf<Int>()

    val edgesCost = mutableListOf<Long>()
    var done = false
    var cost = Long.MAX_VALUE
    var from: Node? = null

    fun reset() {
        done = false
        cost = Long.MAX_VALUE
        from = null
    }
}

private fun calcDijkstraCost(nodes: Map<Int, Node>, start: Int) {
    val queue = PriorityQueue<Node>(11,Comparator<Node> { o1:Node, o2:Node-> (o1.cost - o2.cost).toInt() })

    nodes[start]!!.cost = 0
    queue.add(nodes[start])

    while (queue.isNotEmpty()) {
        val doneNode = queue.remove()
        doneNode.done = true

        for (i in 0 until doneNode.edgesTo.size) {
            val to = doneNode.edgesTo[i]
            val cost = doneNode.cost + doneNode.edgesCost[i]
            if (nodes[to]!!.done.not() && cost < nodes[to]!!.cost) {
                nodes[to]!!.cost = cost
                nodes[to]!!.from = doneNode
                if (queue.contains(nodes[to]).not()) queue.add(nodes[to])
            }
        }
    }
}