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: Wrong Answer)

/**Function Template**/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> pint;
const int MAX = 510000;
const int MOD = 1000000007;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define Rep(i, n) for(ll i = 1; i < (n); i++)
#define ALL(a)  (a).begin(),(a).end()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define PI 3.14159265358979323846
ll fac[MAX], finv[MAX], inv[MAX];


template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

bool palindrome(string s){
  bool flag=true;
  rep(i,s.size()) if(s[i]!=s[s.size()-1-i]) flag=false;
  return flag;
}

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }

}
ll Len(ll n) {
  ll s=0;
  while(n!=0) s++, n/=10;
  return s;
}


ll Sint(ll n) {
  ll m=0,s=0,a=n;
  while(a!=0) s++, a/=10;
  for(ll i=s-1;i>=0;i--) m+=n/((ll)pow(10,i))-(n/((ll)pow(10,i+1)))*10;
  return m;
}


ll Svec(vector<ll> v){
  ll n=0;
  for(ll i=0;i<v.size();i++) n+=v[i];
  return n;
}


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;
}


ll Factorial(ll n){
  ll m=1;
  while(n>=1) m*=n,n--;
  return m;
}

void runlength(string s,vector<pair<char,ll>> &p){
  ll x=1;
  if(s.size()==1){
    p.push_back(pair<char,ll>(s[0],1));
  }
  for(ll i=0;i<s.size()-1;i++){
    if(s[i]==s[i+1]){
      x++;
      if(i==s.size()-2){
        p.push_back(pair<char,ll>(s[i],x));
      }
    }else{
      p.push_back(pair<char,ll>(s[i],x));
      x=1;
      if(i==s.size()-2){
        p.push_back(pair<char,ll>(s[s.size()-1],x));
      }
    }
  }
}

ll COM(ll n,ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

string Toupper(string s){
  string ans="";
  rep(i,s.size()){
    if('a'<=s[i] && s[i]<='z') ans+=(char)s[i]-32;
    else ans+=s[i];
  }
  return ans;
}
string Tolower(string s){
  string ans="";
  rep(i,s.size()){
    if('A'<=s[i] && s[i]<='Z') ans+=(char)s[i]+32;
    else ans+=s[i];
  }
  return ans;
}


const int MAX_N=100010;
vector<bool> sieve_of_eratosthenes(){
  vector<bool> isPrime(MAX_N+1,true);
  for(int i=2;i<=MAX_N;i++){
    if(isPrime[i]){
      for(int j=2*i;j<=MAX_N;j+=i){
        isPrime[j]=false;
      }
    }
  }
  return isPrime;
}

vector<pint> prime_factorize(ll n){
  vector<pint> ans;
  for(ll p=2;p<=sqrt(n);p++){
    if(n%p!=0) continue;
    ll cnt=0;
    while(n%p==0){
      n/=p;
      cnt++;
    }
    ans.push_back(make_pair(p,cnt));
  }
  if(n!=1) ans.push_back(make_pair(n,1));
  return ans;
}

/*bool cmp(pint a, pint b) { return a.second < b.second; }*/
/* ダイクストラ法 */

const int MAX_V=100010;
struct edge{ll to,cost;};
typedef pair<ll,ll> pint;
ll n;
ll INF=1e12;
vector<edge> graph[MAX_V];
ll dist[MAX_V];
void dijkstra(ll s){
  priority_queue<pint,vector<pint>,greater<pint>> que;
  fill(dist,dist+n,INF);
  dist[s]=0;
  que.push(pint(0,s));
  while(!que.empty()){
    auto p=que.top();que.pop();
    ll v=p.second;
    if(dist[v]<p.first) continue;
    for(auto e:graph[v]){
      if(dist[e.to]>dist[v]+e.cost){
        dist[e.to]=dist[v]+e.cost;
        que.push(pint(dist[e.to],e.to));
      }
    }
  }
}
/*nがmain関数に定義されていると動かない*/
/*↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓*/
int main() {
  IOS;
  cin>>n;
  rep(i,n-1){
    ll a,b,c;cin>>a>>b>>c;a--;b--;
    graph[a].push_back(edge{b,c});
    graph[b].push_back(edge{a,c});
  }
  ll q,k;cin>>q>>k;k--;
  dijkstra(k);
  rep(i,q){
    ll s,t;cin>>s>>t;s--;t--;
    cout<<dist[s]+dist[t]<<endl;
  }
}