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 Ruby (2.3.3) (judge verdict: Accepted)

class PriorityQueue

  def initialize(heap=[])
    @size = heap.size
    @heap = heap.sort
  end

  # log( log n )
  def push x
    i = @size
    @size += 1
    while i > 0
      par = ( i - 1 ) / 2
      break if @heap[par] <= x
      @heap[i] = @heap[par]
      i = par
    end
    @heap[i] = x
  end

  # log( log n )
  # return nil if @size == 0
  def pop

    ret = @heap[0]
    x = @heap[@size-=1]


    i = 0
    while (child = i * 2 + 1) < @size
      child_1 = i * 2 + 2
      child = child_1 if child_1 < @size && @heap[child_1] < @heap[child]
      break if @heap[child] >= x
      @heap[i] = @heap[child]
      i = child
    end

    @heap[i] = x
    @heap.delete_at(-1)
    ret
  end

  def size
    @heap.size
  end

  def top
    @heap[0]
  end

  def any?
    @size > 0
  end

  def empty?
    @size == 0
  end

  def all
    @heap
  end

  def first
    @heap[0]
  end

  def second
    @size < 3 ? @heap[1] : [@heap[1], @heap[2]].min
  end

  def third
    @size < 3 ? nil : [@heap[1], @heap[2]].max
  end
end

class Array

  include Comparable

  def to_pq
    PriorityQueue.new(self)
  end
end

class Node

  attr_accessor :to, :cost
  def initialize(to, cost)
    @to = to
    @cost = cost
  end
end

# with PriorityQueue O(|E|log|V|)
def dijkstra(g, s)

  # inf = 10 ** 12
  inf = Float::INFINITY
  dist = Array.new(g.size, inf)
  dist[s] = 0

  que = PriorityQueue.new
  que.push( [0, s] )

  while que.any?

    d, v = que.pop
    next if dist[v] < d
    # p g[v]
    g[v].each do |e|
      if dist[e.to] > dist[v] + e.cost
        dist[e.to] = dist[v] + e.cost
        que.push( [dist[e.to], e.to ] )
      end
    end
    # p que
  end

  # dist.map!{|d| d == inf ? "INF" : d }
  dist
end

n = gets.to_s.to_i
# n = gets.to_s.split.map{|t|t.to_i}

G = Array.new(n){ [] }
(n-1).times do
  s, t, d = gets.to_s.split.map{|t|t.to_i}
  s -= 1
  t -= 1
  G[s] << Node.new(t, d)
  G[t] << Node.new(s, d)
end
# pp G
q, k = gets.to_s.split.map{|t|t.to_i}
k -= 1
dist = dijkstra(G, k)
ans = []
q.times do
  x, y = gets.to_s.split.map{|t|t.to_i-1}
  ans << (dist[x] + dist[y])
end

puts ans


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