Problem:
Score : 500 points

Problem Statement

There are N cities numbered 1 to N, connected by M railroads.

You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.

The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes.
You cannot use gold coins to pay the fare.

There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin.
The transaction takes D_i minutes for each gold coin you give.
You can exchange any number of gold coins at each exchange counter.

For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq 100

0 \leq S \leq 10^9

1 \leq A_i \leq 50

1 \leq B_i,C_i,D_i \leq 10^9

1 \leq U_i < V_i \leq N

There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).

Each city t=2,...,N can be reached from City 1 with some number of railroads.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N

Output

For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.

Sample Input 1

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5

Sample Output 1

2
14

The railway network in this input is shown in the figure below.

In this figure, each city is labeled as follows:

The first line: the ID number i of the city (i for City i)

The second line: C_i / D_i

Similarly, each railroad is labeled as follows:

The first line: the ID number i of the railroad (i for the i-th railroad in input)

The second line: A_i / B_i

You can travel from City 1 to City 2 in 2 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

You can travel from City 1 to City 3 in 14 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 6 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Sample Input 2

4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4

Sample Output 2

5
5
7

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 4 in 7 minutes, as follows:

At the exchange counter in City 1, exchange 2 gold coins for 6 silver coins in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Use the 4-th railroad to move from City 3 to City 4 in 1 minutes.

Sample Input 3

6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1

Sample Output 3

1
9003
14606
16510
16576

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 6 in 16576 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 9000 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

At the exchange counter in City 3, exchange 8 gold coins for 8 silver coins in 5600 minutes.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Use the 3-rd railroad to move from City 2 to City 4 in 1 minute.

At the exchange counter in City 4, exchange 19 gold coins for 19 silver coins in 1900 minutes.

Use the 3-rd railroad to move from City 4 to City 2 in 1 minute.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

Use the 4-th railroad to move from City 3 to City 5 in 1 minute.

At the exchange counter in City 5, exchange 63 gold coins for 63 silver coins in 63 minutes.

Use the 4-th railroad to move from City 5 to City 3 in 1 minute.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 5-th railroad to move from City 1 to City 6 in 1 minute.

Sample Input 4

4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7

Sample Output 4

1
3
5

The railway network in this input is shown in the figure below:

Sample Input 5

2 1 0
1 2 1 1
1 1000000000
1 1

Sample Output 5

1000000001

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 2 in 1000000001 minutes, as follows:

At the exchange counter in City 1, exchange 1 gold coin for 1 silver coin in 1000000000 minutes.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Solution in Scala (2.13.1) (judge verdict: Accepted)

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

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

  case class Edge(to: Int, fee: Int, time: Long)

  case class Node(num: Int, time: Long)

  val N, M, _S = in.next().toInt
  val graph = Array.fill[List[Edge]](N+1)(Nil)
  val nodes = new Array[Node](N+1)
  for (i <- 1 to M) {
    val u, v, a, b = in.next().toInt
    val edge = 
    graph(u) = Edge(v, a, b) :: graph(u)
    graph(v) = Edge(u, a, b) :: graph(v)
  }
  for (i <- 1 to N) {
    val c, d = in.next().toInt
    nodes(i) = Node(c, d)
  }

  val sMax = 2500
  val S = Math.min(_S, sMax)


  //
  val MAX = Long.MaxValue

  val dist = Array.ofDim[Long](N+1, sMax+1) // from からの距離を格納する
  for (i <- 1 to N; j <- 0 to sMax)
    dist(i)(j) = MAX

  val check = Array.ofDim[Boolean](N+1, sMax+1)  // すでにその頂点からを調べたかのフラグ

  dist(1)(S) = 0

  val pq = new PriorityQueue[(Long, (Int, Int))]  // (-cost, node) queue
  pq.enqueue((-0, (S, 1)))

  // 頂点は v 個あるので、v 周する
  for(_ <- 0 until N * 2500) {
    // 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
    var ginka = -1
    var nowTime = -1L

    breakable {
      while(!pq.isEmpty) {
        val (_time, (_ginka, nodeNum)) = pq.dequeue
        nowNode = nodeNum
        ginka = _ginka
        nowTime = -_time
        if(!check(nowNode)(ginka)){  // まだチェックされていないノードだったら決定
          check(nowNode)(ginka) = true
          break
        }
      }
    }

    if (nowNode != -1) {
      val node = nodes(nowNode)
      if (ginka < sMax) {
        val nextTime = nowTime + node.time
        val nextGinka = (ginka + node.num) min sMax
        dist(nowNode)(nextGinka) = dist(nowNode)(nextGinka) min nextTime

        pq.enqueue((-nextTime, (nextGinka, nowNode)))
      }

      // その頂点からたどり着ける頂点の情報を更新する
      for(Edge(nextNode, fee, time) <- graph(nowNode)) {
        val nextTime = nowTime + time
        val nextGinka = ginka - fee

        if(nextGinka >= 0 && nextTime < dist(nextNode)(nextGinka)){  // 既存の通路より近いなら更新
          dist(nextNode)(nextGinka) = nextTime
          pq.enqueue((-nextTime, (nextGinka, nextNode)))
        }
      }
    }
  }

  for (i <- 2 to N) {
    pw.println(dist(i).min)
  }
  pw.flush()
}

object Sub {
  def print2DimArray[A](mat: Array[Array[A]]): Unit = {
    mat.foreach(lin => println(lin.toList))
  }
}

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