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
stdin = sys.stdin

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

from heapq import heappush, heappop

def dijkstra(graph:list, node:int, start:int) -> list:
    # 未探索のノードは距離INF
    INF = float("inf")
    dist = [INF]*node
    
    # 始点ノードの距離を0とし、dfsのためのpriority queを作成
    dist[start] = 0
    heap = [(0,start)]
    
    # 未探索のノードをpriority queueに入れる
    while heap:
        cost, cur_node = heappop(heap)
        
        for nex_cost, nex_node in graph[cur_node]:
            dist_cand = dist[cur_node] + nex_cost
            if dist_cand < dist[nex_node]:
                dist[nex_node] = dist_cand
                heappush(heap, (dist[nex_node], nex_node))
    
    return dist

n = ni()
abc = []
for _ in range(n-1):
    a,b,c = li()
    a -= 1
    b -= 1
    abc.append((a,b,c))
q,k = li()
k -= 1
query = [li_() for _ in range(q)]

# 隣接リストつくる
graph = [[] for _ in range(n)]
for frm,to,cost in abc:
    graph[frm].append((cost,to))
    graph[to].append((cost,frm))

# 頂点Kをスタートとするダイクストラ
dist = dijkstra(graph, n, k)

# クエリ処理
for frm,to in query:
    print(dist[frm] + dist[to])