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 Python (3.4.3) (judge verdict: Accepted)

import sys
input = sys.stdin.readline

n = int(input())
adj = [[] for _ in range(n)]  # adjacence(隣接)の略

for (a,b,c) in (map(int, input().split()) for _ in range(n-1)):
    adj[a-1].append((b-1, c))
    adj[b-1].append((a-1, c))

q, k = map(int, input().split())
k -= 1
dist = [-1]*n  # 経由地から各頂点までのコスト
dist[k] = 0  # 経由地kをスタート地点とするため、コスト0に設定

stack = [k]  # 探索候補。経由地kをスタート地点にするため、先に積んでおく

while stack: # 探索候補がなくなるまで続ける
    now = stack.pop()
    for dest, cost in adj[now]:
        if dist[dest] != -1: # dest(目的地)までのコストがすでに設定されていればスルー
            continue
        dist[dest] = dist[now] + cost # コストを現地点までのコストと合算して算出
        stack.append(dest)

for _ in range(q):
    x, y = map(int, input().split())
    print(dist[x-1] + dist[y-1])