Problem:
Score : 400 points

Problem Statement

You are given a tree with N vertices.

Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N-1 edges, where N is the number of its vertices.

The i-th edge (1≤i≤N-1) connects Vertices a_i and b_i, and has a length of c_i.

You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):

find the length of the shortest path from Vertex x_j and Vertex y_j via Vertex K.

Constraints

3≤N≤10^5

1≤a_i,b_i≤N (1≤i≤N-1)

1≤c_i≤10^9 (1≤i≤N-1)

The given graph is a tree.

1≤Q≤10^5

1≤K≤N

1≤x_j,y_j≤N (1≤j≤Q)

x_j≠y_j (1≤j≤Q)

x_j≠K,y_j≠K (1≤j≤Q)

Input

Input is given from Standard Input in the following format:

N
a_1 b_1 c_1
:
a_{N-1} b_{N-1} c_{N-1}
Q K
x_1 y_1
:
x_{Q} y_{Q}

Output

Print the responses to the queries in Q lines.

In the j-th line j(1≤j≤Q), print the response to the j-th query.

Sample Input 1

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

Sample Output 1

3
2
4

The shortest paths for the three queries are as follows:

Query 1: Vertex 2 → Vertex 1 → Vertex 2 → Vertex 4 : Length 1+1+1=3

Query 2: Vertex 2 → Vertex 1 → Vertex 3 : Length 1+1=2

Query 3: Vertex 4 → Vertex 2 → Vertex 1 → Vertex 3 → Vertex 5 : Length 1+1+1+1=4

Sample Input 2

7
1 2 1
1 3 3
1 4 5
1 5 7
1 6 9
1 7 11
3 2
1 3
4 5
6 7

Sample Output 2

5
14
22

The path for each query must pass Vertex K=2.

Sample Input 3

10
1 2 1000000000
2 3 1000000000
3 4 1000000000
4 5 1000000000
5 6 1000000000
6 7 1000000000
7 8 1000000000
8 9 1000000000
9 10 1000000000
1 1
9 10

Sample Output 3

17000000000

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

import scala.collection.mutable

object Main {

  object Scanner {
    private val buf = new Array[Byte](1024); private var ptr = 0; private var len = 0
    @inline private def isPrintableChar(c: Int): Boolean = 33 <= c && c <= 126
    @inline private def hasNextByte(): Boolean =
      if (ptr >= len) { ptr = 0; len = System.in.read(buf); len > 0 } else { true }
    @inline private def hasNext(): Boolean = {
      while (hasNextByte() && !isPrintableChar(buf(ptr))) ptr += 1
      hasNextByte()
    }
    @inline private def readByte(): Byte =
      if (hasNextByte()) { val res = buf(ptr); ptr += 1; res } else { -1 }
    def next(): String = {
      if(!hasNext()) ???
      val sb = new StringBuilder; var b = readByte()
      while (isPrintableChar(b)) { sb.append(b.toChar); b = readByte() }
      sb.toString
    }
    def nextInt(): Int = {
      val n = nextLong()
      if (n < Int.MinValue || Int.MaxValue < n) ???
      n.toInt
    }
    def nextLong(): Long = {
      if(!hasNext()) ???
      var minus = false; var b = readByte()
      if (b == '-') { minus = true; b = readByte() }
      def go (b: Byte, n: Long = 0): Long =
        if ('0' <= b && b <= '9') { go(readByte(), n * 10 + b - '0') }
        else if (minus) { -n } else { n }
      go(b)
    }
    def nextDouble(): Double = next.toDouble
  }

  case class Edge(v: Int, w: Long)

  def main(args: Array[String]): Unit = {
    val sc = Scanner
    val INF = Long.MaxValue
    val V = sc.nextInt()
    val G = Array.fill(V + 1)(List[Edge]())
    (0 until V - 1) foreach { _ =>
      val u, v, w = sc.nextInt()
      G(u) ::= Edge(v, w)
      G(v) ::= Edge(u, w)
    }
    val Q, K = sc.nextInt()
    val X, Y = new Array[Int](Q)
    (0 until Q) foreach { i =>
      val x, y = sc.nextInt()
      X(i) = x
      Y(i) = y
    }

    // j番目の質問クエリ、頂点xjから頂点Kを経由しつつ、頂点yjまで
    // 移動する場合の最短経路の距離を求めてください
    // ...つまり, xjからKへの最短距離, Kからyjへの最短距離の和を答えればそれが最短経路となる
    // 木に対する全点対最短路問題を事前に解いておき, クエリへ回答する
    // 木かつ ci >= 1 なので, ダイクストラ法によって全点対最短路を求めることができる.

    def dijkstra(s: Int, V: Int, E: Int, G: Array[List[Edge]]): Array[Long] = {
      val d = Array.fill[Long](V+1)(INF)
      val queue = mutable.PriorityQueue[(Long, Int)]()(Ordering.by(_._1))
      d(s) = 0L
      queue.enqueue((0L, s))

      def loop(): Unit =
        if (queue.nonEmpty) {
          val (ud, u) = queue.dequeue()
          if (d(u) < ud) loop()
          else {
            d(u) = ud
            G(u) foreach { edge =>
              if (d(edge.v) > d(u) + edge.w) {
                d(edge.v) = d(u) + edge.w
                queue.enqueue((d(edge.v), edge.v))
              }
            }
            loop()
          }
        }
      loop()
      d
    }

    val d: Array[Array[Long]] =
      Array.tabulate(V)(s => dijkstra(s + 1, V, V - 1, G))

    val r = (0 until Q).map { i =>
      d(X(i)-1)(K) + d(K-1)(Y(i))
    }. mkString("\n")

    println(r)
  }

}