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: Accepted)

#include <bits/stdc++.h>
using namespace std;

// エイリアス
using  ll = long signed long;
using ull = long unsigned long;
using  ld = long double;
using   P = pair<int, int>;
using llP = pair<ll, ll>;
using DoP = pair<double, double>;

// 汎用マクロ
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define rep(i, n) for (int i = 0; i < n; ++i)
#define FOREACH(i,q) for (auto &i : q)
#define UNIQUE(v) do { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); } while (false)
#define SZ(x) ((int)(x).size())
template <class T>inline bool chmin(T &a, const T b) {if (a > b) { a = b; return true;} return false;}
template <class T>inline bool chmax(T &a, const T b) {if (a < b) { a = b; return true;} return false;}
template <class T>inline void line_out(const vector<T> vec) {int n = SZ(vec); rep(i, n) { cout << vec[i]; if(i < n-1) cout << " ";}cout << endl;}
const int di[] = {0, 1, 0, -1, 1, 1, -1, -1};
const int dj[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int INF = 1 << 28;
const ll INF64 = 1ll << 60;
const int mod = 1000000007;
//const int mod = 998244353;
struct mint
{
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a)
  {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a)
  {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a)
  {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const
  {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const
  {
    mint res(*this);
    return res -= a;
  }
  mint &operator++()
  {
    if ((x += 1) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator--()
  {
    if ((x += mod - 1) >= mod)
      x -= mod;
    return *this;
  }
  mint operator*(const mint a) const
  {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const
  {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const
  {
    return pow(mod - 2);
  }
  mint &operator/=(const mint a)
  {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const
  {
    mint res(*this);
    return res /= a;
  }
};

struct edge{int to; ll cost;};

int main()
{
  int n; cin >> n;
  vector<vector<edge>> g(n+1);
  rep(i, n-1) {
    int a, b;
    ll c;
    cin >> a >> b >> c;
    g[b].push_back(edge{a, c});
    g[a].push_back(edge{b, c});
  }
  int q, k; cin >> q >> k;
  vector<P> query(q);
  rep(i, q) cin >> query[i].first >> query[i].second;
  vector<ll> dist(n+1, INF64);
  priority_queue<llP, vector<llP>, greater<llP>> pq;
  dist[k] = 0;
  pq.push(llP(0, k));

  while(!pq.empty()) {
    auto p = pq.top(); pq.pop();
    int u = p.second;
    ll cost = p.first;

    if(dist[u] < cost) continue;

    for(auto next : g[u]) {
      int v = next.to;
      if(chmin(dist[v], cost+next.cost)) {
        pq.push(llP(dist[v], v));
      }
    }
  }
  rep(i, q) {
    cout << dist[query[i].first]+dist[query[i].second] << endl;
  }
}