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<iomanip>
#include<math.h>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<string>
#define INF 1000000000ll
#define MOD 1000000007ll
#define EPS 1e-8
#define REP(i,m) for(long long i=0; i<m; ++i)
#define FOR(i,n,m) for(long long i=n; i<m; ++i)
#define ALL(v) v.begin(),v.end()
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vector<int> > vii;
typedef vector<long long> vl;
typedef vector<vector<long long> > vll;
typedef vector<P> vp;
typedef vector<vector<P> > vpp;
struct Edge {
	int from;
	int to;
	ll cost;
};
// make the adjacent list from the edge list
vector<vector<P> > makeAdj(int k, int n, const vector<Edge>& e) {
	// n:the number of vertex
	// k=0:undirected, k=1:directed
	vector<vector<P> > ret(n, vector<P>(0));
	for(int i=0; i<e.size(); ++i) {
		ret[e[i].from].push_back(P(e[i].to, e[i].cost));
		if(k==0) ret[e[i].to].push_back(P(e[i].from, e[i].cost));
	}
	return ret;
}
// Do dijkstra
vector<ll> dijkstra(int s, const vector<vector<P> >& adj) {
	int n=(int)adj.size();
	vector<ll> dis(n, INF*INF);
	priority_queue<P, vector<P>, greater<P> > q;
	dis[s]=0;
	q.push(P(0, s));
	while(!q.empty()) {
		P buf=q.top();
		q.pop();
		ll d=buf.first;
		ll p=buf.second;
		if(dis[p]<d) continue;
		for(int i=0; i<adj[p].size(); ++i) {
			ll nd=adj[p][i].second;
			ll np=adj[p][i].first;
			if(dis[np]>dis[p]+nd) {
				dis[np]=dis[p]+nd;
				q.push(P(dis[np], np));
			}
		}
	}
	return dis;
}int main() {
	ios::sync_with_stdio(false);
	int N;
	cin>>N;
	vector<Edge> e(N-1);
	REP(i,N-1) {
		int a,b,c;
		cin>>a>>b>>c;
		--a;
		--b;
		e[i].from=a;
		e[i].to=b;
		e[i].cost=c;
	}
	vector<vector<P> > adj=makeAdj(0,N,e);
	int q,k;
	cin>>q>>k;
	--k;
	vector<ll> dis=dijkstra(k,adj);
	REP(i,q) {
		int x,y;
		cin>>x>>y;
		--x;
		--y;
		cout<<dis[x]+dis[y]<<endl;
	}
}