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++ (GCC 9.2.1) (judge verdict: Memory Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using V = vector<int>;
using VV = vector<V>;
using VVV = vector<VV>;
using VL = vector<ll>;
using VVL = vector<VL>;
using VVVL = vector<VVL>;
template<class T> using VE = vector<T>;
template<class T> using P = pair<T, T>;
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define REP(i,k,n) for(int i=(k);i<(n);i++)
#define all(a) (a).begin(),(a).end()
#define output(x,y) cout << fixed << setprecision(y) << x << endl;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
const ll MOD = 1e9 + 7;
// const ll MOD = 998244353;
ll upper = MOD + MOD;
ll under = -upper;
ll UPPER = MOD * MOD;
ll UNDER = -UPPER;
const long double pi = 3.141592653589793;
const int N = 101010;
VL dist(N, UPPER);
void dfs(int x, VE<VE<P<int>>>& G,ll Dist,VE<bool> visited) { // 実は木上の最短経路はdfsで行ける
	visited[x] = true;
	dist[x] = Dist;
	for (auto c : G[x]) {
		if (!visited[c.first]) dfs(c.first, G, Dist + c.second, visited);
	}
}
int main() {
	int n;
	cin >> n;
	VE<VE<P<int>>> G(n);
	rep(i, n - 1) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		a--, b--;
		G[a].push_back(make_pair(b, c));
		G[b].push_back(make_pair(a, c));
	}
	int q, k;
	cin >> q >> k;
	dist[--k] = 0;
	VE<bool> visited(N, false);
	dfs(k, G, 0, visited);
	while (q--) {
		int x, y;
		scanf("%d%d", &x, &y);
		x--, y--;
		printf("%lld\n", dist[x] + dist[y]);
	}
	return 0;
}