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>
using namespace std;
typedef long long ll;
const ll N = 4e5+9;
ll n,k,q,tdfn,tot;
ll dp[50][N],vis[N];
ll B[N],LOG[N],used[N],F[N],pos[N];
vector<pair<ll,ll> > edge[N];
void rmq_init(ll n,ll num[])
{
    ll i,j;
    for(j=1;j<=n;j++) 
        dp[0][j]=num[j];
    for(j=1;j<=LOG[n];j++)
    {
        ll limit=n+1-(1<<j);
        for(i=1;i<=limit;i++)
        {
            ll x=i+(1<<j>>1);
            dp[j][i]=min(dp[j-1][x],dp[j-1][i]);
        }
    }
}
ll rmq(ll l,ll r,ll num[])
{
    ll m=LOG[r-l+1];
    return min(dp[m][l],dp[m][r-(1<<m)+1]);
}
ll sum[N];
void dfs(ll s)
{
    ll i,t;
    used[s]=1;
    ll tmp=++tdfn;
    B[++tot]=tmp;F[tmp]=s;
    pos[s]=tot;
    for(i=0;i<edge[s].size();i++)
    {
        t=edge[s][i].first;
        if(used[t]) continue;
        sum[t]=sum[s]+edge[s][i].second;
        dfs(t);
        B[++tot]=tmp;
    }
}
ll lca(ll a,ll b)
{
    if(pos[a]>pos[b]) swap(a,b);
    ll ans=rmq(pos[a],pos[b],B);
    return F[ans];
}
int main()
{
    ll i,a,b,w;
    LOG[0]=-1;
    for(i=1;i<N;i++)  LOG[i]=LOG[i>>1]+1;
    cin >> n;
    for(i=0;i<=n;i++) edge[i].clear();
    for(i=1;i<n;i++)
    {
        cin >> a >> b >> w;
        edge[a].push_back(make_pair(b,w));
        edge[b].push_back(make_pair(a,w));
    }
    sum[1]=0;
    tdfn=0;
    tot=0;
    memset(used,0,sizeof(used));
    dfs(1);
    rmq_init(tot,B);
    cin >> q >> k;
    while(q--)
    {
        cin >> a >> b;
        cout << sum[a]+sum[k]-2*sum[lca(a,k)] + sum[b]+sum[k]-2*sum[lca(b,k)] << endl;
    }
    return 0;
}