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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>

using IV = std::vector<int>;
IV ToFixedIntArray(std::istream& is, int n) {
	IV ret(n);
	auto* dst = ret.data();
	while(n > 0) {
		--n;
		is >> *dst++;
	}
	return ret;
}

#define _For(idx, from, to, inc, cmp) \
	for(auto idx=(from) ; idx cmp (to) ; inc idx)
#define ForLE(idx, from, to) \
	_For(idx, from, to, ++, <=)
#define ForGE(idx, from, to) \
	_For(idx, from, to, --, >=)
#define ForLT(idx, from, to) \
	_For(idx, from, to, ++, <)
#define ForGT(idx, from, to) \
	_For(idx, from, to, --, >)
#define For(idx, from, to) \
	_For(idx, from, to, ++, <)
#define Assert(expr) if(!(expr)) {__builtin_trap();}

void DisableIOSync() {
	std::cin.tie(0);
	std::cout.tie(0);
	std::ios::sync_with_stdio(false);
}
struct Node {
	using Link = std::vector<std::pair<Node*, int>>;
	Link	link;
	uint64_t toKCost;
};

#include <unordered_set>
std::unordered_set<Node*> Used;
void CalcCost(Node* node, const uint64_t cost) {
	if(Used.count(node) != 0)
		return;
	Used.emplace(node);
	node->toKCost = cost;
	for(auto& nd : node->link) {
		CalcCost(nd.first, cost + nd.second);
	}
}
int main() {
	DisableIOSync();

	const int N = ToFixedIntArray(std::cin, 1)[0];
	std::vector<Node> node(N);
	For(i, 0, N-1) {
		const auto inp = ToFixedIntArray(std::cin, 3);
		const int i0 = inp[0]-1,
				i1 = inp[1]-1;
		const int cost = inp[2];
		node[i0].link.emplace_back(&node[i1], cost);
		node[i1].link.emplace_back(&node[i0], cost);
	}
	const auto qk = ToFixedIntArray(std::cin, 2);
	const auto Q = qk[0],
				K = qk[1]-1;

	CalcCost(&node[K], 0);
	For(i, 0, Q) {
		const auto query = ToFixedIntArray(std::cin, 2);
		std::cout << node[query[0]-1].toKCost + node[query[1]-1].toKCost << "\n";
	}
	std::cout.flush();
	return 0;
}