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 <bits/stdc++.h>

#define ALL(v) (v).begin(), (v).end()
#define rep(i, n) for (ll i = 0, __##i##_length = (n); i < __##i##_length; i++)
#define rep_reverse(i, n) for (ll i = (n)-1; 0 <= i; i--)
#define FOR(i, from, to) for (ll i = (from), __##i##_end = (to); i < __##i##_end; i++)
#define FOR_REVERSE(i, from, to) for (ll i = (to)-1, __##i##_end = (from); __##i##_end <= i; i--)
#define foreach(e, v) for (const auto e : (v))
#define vfree(v) std::vector<int>().swap(v)
#define DEBUG(x) cout << #x << " = " << (x) << endl

using namespace std;
typedef int64_t ll;
typedef vector<ll> llv;
typedef vector<llv> llvv;
constexpr ll ll_max = numeric_limits<ll>::max();

void dfs(vector<llvv>& G, llv& mindist, ll x)
{
  rep(i, G[x].size())
  {
    llv t = G[x][i];
    if (mindist[t[0]] != -1)
      continue;
    ll a = mindist[x];
    ll b = t[1];
    mindist[t[0]] = a + b;
    dfs(G, mindist, t[0]);
  }
}

int main()
{
  ll N, Q, K;
  cin >> N;
  vector<llvv> G(N);

  rep(i, N - 1)
  {
    ll a, b, c;
    cin >> a >> b >> c;
    G[a - 1].push_back(llv({b - 1, c}));
    G[b - 1].push_back(llv({a - 1, c}));
  }

  llv mindist(N, -1);
  cin >> Q >> K;
  mindist[K - 1] = 0;

  dfs(G, mindist, K - 1);

  rep(i, Q)
  {
    ll x, y;
    cin >> x >> y;
    cout << mindist[x-1] + mindist[y-1] << endl;
  }

  return 0;
}