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: Time Limit Exceeded)

N = gets.to_i
abcs = (N - 1).times.map { gets.split.map(&:to_i) }
Q, K = gets.split.map(&:to_i)
xys = Q.times.map { gets.split.map(&:to_i) }

graph = Array.new(N) { [] }
abcs.each do |(a, b, c)|
  a -= 1
  b -= 1
  graph[a].push([b, c])
  graph[b].push([a, c])
end

def bfs(graph, max_k)
  parent_list = Array.new(max_k + 1) { Array.new(N) }
  cost_list = Array.new(N, 0)
  depth_list = Array.new(N)
  # v, cost, parent, depth
  queue = [[0, 0, nil, 0]]
  while !queue.empty?
    v, cost, parent, depth = queue.shift
    cost_list[v] = cost
    parent_list[0][v] = parent
    depth_list[v] = depth
    graph[v].each do |(u, c)|
      next if u == parent
      queue.push([u, cost + c, v, depth + 1])
    end
  end
  [cost_list, parent_list, depth_list]
end

def lca(u, v, parent_list, depth_list, max_k)
  d1 = depth_list[u]
  d2 = depth_list[v]
  u, v = d1 > d2 ? [v, u] : [u, v]
  # lca
  (0..max_k).each do |k|
    depth_diff = depth_list[v] - depth_list[u]
    if depth_diff[k] == 1
      v = parent_list[k][v]
    end
  end
  return u if u == v
  max_k.downto(0) do |k|
    if parent_list[k][u] != parent_list[k][v]
      u = parent_list[k][u]
      v = parent_list[k][v]
    end
  end
  parent_list[0][u]
end

max_k = Math.log2(N).to_i
cost_list, parent_list, depth_list = bfs(graph, max_k)
(0..(max_k - 1)).each do |k|
  N.times do |v|
    parent_list[k + 1][v] = if parent_list[k][v]
      parent_list[k][parent_list[k][v]]
    else
      nil
    end
  end
end
xys.each do |(x, y)|
  x -= 1
  y -= 1
  x_parent = lca(x, K - 1, parent_list, depth_list, max_k)
  k_parent = lca(K - 1, y, parent_list, depth_list, max_k)
  cost =
    (cost_list[x] - cost_list[x_parent]).abs + # x -> x_parent
    (cost_list[x_parent] - cost_list[K - 1]).abs + # x_parent -> k
    (cost_list[K - 1] - cost_list[k_parent]).abs + # k -> k_parent
    (cost_list[k_parent] - cost_list[y]).abs # k_parent -> y
  puts cost
end