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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.stream.IntStream;

/**
 * 解説どおりに作成したソースコード
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			@SuppressWarnings("unchecked")
			List<Path>[] pathLists = new List[n];
			IntStream.range(0, n).forEach(i -> pathLists[i] = new ArrayList<>());
			IntStream.range(0, n - 1).forEach(i -> {
				int a = scanner.nextInt() - 1;
				int b = scanner.nextInt() - 1;
				long c = scanner.nextLong();
				pathLists[a].add(new Path(a, b, c));
				pathLists[b].add(new Path(b, a, c));
			});
			int q = scanner.nextInt(), k = scanner.nextInt() - 1;
			long[] distances = dijkstra(pathLists, k, n);
			IntStream.range(0, q).forEach(
					i -> System.out.println(distances[scanner.nextInt() - 1] + distances[scanner.nextInt() - 1]));
		}
	}

	/**
	 * @param pathLists ノード間の連結パス一覧の配列
	 * @param src       開始ノード
	 * @param n         ノードの数
	 * @return 開始ノードからの距離の配列
	 */
	private static long[] dijkstra(List<Path>[] pathLists, int src, int n) {
		long[] distances = new long[n];
		Arrays.fill(distances, Long.MAX_VALUE);
		// 最短距離が確定したかどうか
		boolean[] fixed = new boolean[n];
		Arrays.fill(fixed, false);
		// 出発地点までの距離を0とする
		distances[src] = 0L;
		Queue<Path> queue = new PriorityQueue<>();
		queue.add(new Path(src, src, 0));
		while (!queue.isEmpty()) {
			int now = queue.poll().to;
			if (!fixed[now]) {
				fixed[now] = true;
				for (Path path : pathLists[now]) {
					if (!fixed[path.to] && distances[path.to] > distances[now] + path.cost) {
						distances[path.to] = distances[now] + path.cost;
						queue.add(new Path(path.to, path.to, distances[path.to]));
					}
				}
			}
		}
		return distances;
	}

	/**
	 * ノード間の一つのパスを表すクラス
	 */
	private static class Path implements Comparable<Path> {
		/** 開始ノード */
		@SuppressWarnings("unused")
		int from;
		/** 終了ノード */
		int to;
		/** 移動コスト */
		long cost;

		Path(int from, int to, long cost) {
			super();
			this.from = from;
			this.to = to;
			this.cost = cost;
		}

		@Override
		public int compareTo(Path path) {
			if (this.cost > path.cost) {
				return 1;
			} else if (this.cost == path.cost) {
				return 0;
			} else {
				return -1;
			}
		}
	}
}