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 PyPy3 (7.3.0) (judge verdict: Accepted)

from collections import deque


def nearlist(N, LIST):  # 隣接リスト
    NEAR = [set() for _ in range(N)]
    COST = {}
    for a, b, c in LIST:
        NEAR[a - 1].add(b - 1)
        NEAR[b - 1].add(a - 1)
        COST[(a - 1, b - 1)] = c
        COST[(b - 1, a - 1)] = c
    return NEAR, COST


def bfs(NEAR, S, N):  # 幅優先探索  # キュー
    DIST = [-1 for _ in range(N)]  # 前処理
    DIST[S] = 0
    que, frag = deque([S]), set([S])

    while len(que) > 0:
        q = que.popleft()
        for i in NEAR[q]:  # 移動先の候補
            if i in frag:  # 処理済みか否か
                continue
            DIST[i] = DIST[q] + cost[(q, i)]
            que.append(i), frag.add(i)
    return DIST


n = int(input())
abc = [list(map(int, input().split())) for _ in range(n - 1)]
q, k = map(int, input().split())
xy = [list(map(int, input().split())) for _ in range(q)]

near, cost = nearlist(n, abc)
dist = bfs(near, k - 1, n)
for x, y in xy:
    print(dist[x - 1] + dist[y - 1])