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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    static long INF = 1001001001001001001L;
    
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        
        // to: 隣接リスト(有向)
        List<int[]>[] to = new List[n];
        for (int i = 0; i < n; i++) {
            to[i] = new ArrayList<int[]>();
        }
        for (int i = 0; i < n-1; i++) {
            int p = sc.nextInt() - 1;
            int q = sc.nextInt() - 1;
            int dis = sc.nextInt();
            to[p].add(new int[]{q, dis});
            to[q].add(new int[]{p, dis});
        }
        
        int query = sc.nextInt();
        int start = sc.nextInt()-1;
        
        // dist: startからの最短距離
        long[] dist = new long[n];
        Arrays.fill(dist, INF);
        dist[start] = 0;
        
        // ダイクストラ法: BFSで距離を配る。
        // (キューの先頭の点はそう来る他最短経路がないので最短距離が確定する)
        PriorityQueue<Point> que = new PriorityQueue<>();
        que.add(new Point(start, 0));
        while (!que.isEmpty()) {
            Point cur = que.poll();
            int p = cur.name;
            long dis = (long)cur.dist;
            
            // distより大きいdisを持つ点からは配らせない
            if (dis > dist[p]) continue;
            
            // 子に「自分の距離 + p→qの距離」を配る
            for (int[] qData : to[p]) {
                int q = qData[0];
                int PtoQ = qData[1];
                
                // 最短距離が更新される場合だけキューに入れる
                if (dis + PtoQ < dist[q]) {
                    dist[q] = dis + PtoQ;
                    que.add(new Point(q, dist[q]));
                }
            }
        }
        
        while (query-- > 0) {
            int px = sc.nextInt()-1;
            int qx = sc.nextInt()-1;
            System.out.println(dist[px]+dist[qx]);
        }
    }
}
    
class Point implements Comparable<Point> {
    int name;
    long dist;
     
    Point(int name, long dist) {
        this.name = name;
        this.dist = dist;
    }
    
    @Override
    public int compareTo(Point o) {
        //重みの小さい順
        if (this.dist < o.dist) {
            return -1;
        } 
        return 1;
    }
}