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 (2.4.0) (judge verdict: Accepted)

import sys
input = sys.stdin.readline

def make_Graph(N, edges):#頂点の名前はi - 1で受け付ける
    graph = [ [] for i in range(N)]
    for edge in edges:
        a, b, c = edge
        graph[a].append((b,c))
        graph[b].append((a,c))
    return graph

def BFS(N, graph, start):
    Q = []
    V = []
    prev = [None] * N
    dd = [float("inf")] * N
    Q.append(start)
    dd[start] = 0
    while len(V) < N:
        a = Q.pop(0)
        V.append(a)
        for b, c in graph[a]:
            if b != prev[a]:
                Q.append(b)
                prev[b] = a
                dd[b] = dd[a] + c
    return dd

def DFS(N, graph, a, p, d):
    dd[a] = d
    for b, c in graph[a]:
        if b == p or dd[b] != 0:
            continue
        else:
            DFS(N, graph, b, a, d + c)

N = int(input())

sys.setrecursionlimit(100)

graph = [ [] for i in range(N)]
for i in range(N-1):
    a, b, c = (int(j) for j in input().split())
    graph[a-1].append((b-1,c))
    graph[b-1].append((a-1,c))

Q, K =(int(i) for i in input().split())
dd = [0] * N

try:
    DFS(N, graph, K-1, -1, 0)
except:
    dd = BFS(N, graph, K-1)

for i in range(Q):
    x, y = (int(a) for a in input().split())
    print(dd[x-1] + dd[y-1])