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 <sstream>
#include <vector>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iterator>
#include <limits>
#include <numeric>
#include <utility>
#include <type_traits>
#include <cmath>
#include <cassert>
#include <cstdio>

using namespace std;
using namespace placeholders;

using LL = long long;
using ULL = unsigned long long;
using VI = vector< int >;
using VVI = vector< vector< int > >;
using VS = vector< string >;
using ISS = istringstream;
using OSS = ostringstream;
using PII = pair< int, int >;
using VPII = vector< pair< int, int > >;
template < typename T = int > using VT = vector< T >;
template < typename T = int > using VVT = vector< vector< T > >;
template < typename T = int > using LIM = numeric_limits< T >;

template < typename T > inline istream& operator>>( istream &s, vector< T > &v ){ for ( T &t : v ) { s >> t; } return s; }
template < typename T > inline ostream& operator<<( ostream &s, const vector< T > &v ){ for ( int i = 0; i < int( v.size() ); ++i ){ s << ( " " + !i ) << v[i]; } return s; }

void in_impl(){};
template < typename T, typename... TS > void in_impl( T &head, TS &... tail ){ cin >> head; in_impl( tail ... ); }
#define IN( T, ... ) T __VA_ARGS__; in_impl( __VA_ARGS__ );

template < typename T > struct getv_fmt;
template <> struct getv_fmt<       int >{ static constexpr const char *fmt = "%d"; };
template <> struct getv_fmt< long long >{ static constexpr const char *fmt = "%lld"; };
template < typename T > void getv( std::vector< T > &v ){ for_each( begin( v ), end( v ), []( T &a ){ scanf( getv_fmt< T >::fmt, &a ); } ); };

template < typename T > inline T fromString( const string &s ) { T res; istringstream iss( s ); iss >> res; return res; }
template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); }

#define NUMBERED( name, number ) NUMBERED2( name, number )
#define NUMBERED2( name, number ) name ## _ ## number
#define REP1( n ) REP2( NUMBERED( REP_COUNTER, __LINE__ ), n )
#define REP2( i, n ) REP3( i, 0, n )
#define REP3( i, m, n ) for ( int i = ( int )( m ); i < ( int )( n ); ++i )
#define GET_REP( a, b, c, F, ... ) F
#define REP( ... ) GET_REP( __VA_ARGS__, REP3, REP2, REP1 )( __VA_ARGS__ )
#define FOR( e, c ) for ( auto &&e : c )
#define ALL( c ) begin( c ), end( c )
#define AALL( a ) ( remove_all_extents< decltype( a ) >::type * )a, ( remove_all_extents< decltype( a ) >::type * )a + sizeof( a ) / sizeof( remove_all_extents< decltype( a ) >::type )
#define DRANGE( c, p ) begin( c ), begin( c ) + ( p ), end( c )

#define SZ( v ) ( (int)( v ).size() )
#define EXIST( c, e ) ( ( c ).find( e ) != ( c ).end() )

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

#define PB push_back
#define EM emplace
#define EB emplace_back
#define BI back_inserter

#define MP make_pair
#define fst first
#define snd second

#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

// 木の根を適当に固定して、最近共通祖先を求める
// 二点間距離もいける
// 前処理 O( N log N )
// クエリ O( log N )
class LowestCommonAncestor
{
private:
	const int N;
	std::vector< VPII > G;

	bool constructed = false;
	std::vector<LL> depth_, distances_;
	std::vector< std::vector<int> > parent_;

public:
	LowestCommonAncestor( const int n ) : N( n ), G( N ), depth_( N ), distances_( N ), parent_( N, std::vector<int>( 31, -1 ) )
	{
		return;
	}

	void connect( const int u, const int v, const int d = 1 )
	{
		G[u].EB( v, d );
		G[v].EB( u, d );
		constructed = false;

		return;
	}

	int depth( const int u ) const
	{
		return depth_[u];
	}

	int solve( int u, int v )
	{
		if ( !constructed )
		{
			dfs( 0, -1, 0, 0 );
			doubling();
			constructed = true;
		}

		if ( depth_[u] < depth_[v] )
		{
			swap( u, v );
		}

		{ // align depths
			int diff = depth_[u] - depth_[v];
			for ( int i = 30; 0 <= i; --i )
			{
				if ( diff & 1 << i )
				{
					u = parent_[u][i];
				}
			}
		}

		if ( u == v )
		{
			return u;
		}

		for ( int i = 30; 0 <= i; --i )
		{
			if ( parent_[u][i] != parent_[v][i] )
			{
				u = parent_[u][i];
				v = parent_[v][i];
			}
		}

		return parent_[u][0];
	}

	LL distance( const int u, const int v )
	{
		const int w = solve( u, v );
		return distances_[u] + distances_[v] - 2 * distances_[w];
	}

private:
	void dfs( const int u, const int prev, const int d, const LL dist )
	{
		depth_[u] = d;
		parent_[u][0] = prev;
		distances_[u] = dist;

		for ( const PII &e : G[u] )
		{
			const int v = e.fst;
			if ( v == prev )
			{
				continue;
			}

			dfs( v, u, d + 1, dist + e.snd );
		}

		return;
	}

	void doubling()
	{
		for ( int i = 0; i < 30; ++i )
		{
			for ( int u = 0; u < N; ++u )
			{
				if ( parent_[u][i] != -1 )
				{
					parent_[u][ i + 1 ] = parent_[ parent_[u][i] ][i];
				}
			}
		}
		return;
	}
};
// LowestCommonAncestor( N )
// connect( u, v [, dist = 1 ] )
// solve( u, v )
// depth( u )
// distance( u, v )

int main()
{
	cin.tie( 0 );
	ios::sync_with_stdio( false );
	cout << setprecision( 12 ) << fixed;

	int N;
	scanf( "%d", &N );

	LowestCommonAncestor lca( N );
	REP( N - 1 )
	{
		int a, b, c;
		scanf( "%d%d%d", &a, &b, &c );
		--a, --b;

		lca.connect( a, b, c );
	}

	int Q, K;
	scanf( "%d%d", &Q, &K );
	--K;

	REP( Q )
	{
		int x, y;
		scanf( "%d%d", &x, &y );
		--x, --y;

		printf( "%lld\n", lca.distance( x, K ) + lca.distance( K, y ) );
	}

	fflush( stdout );
	

	return 0;
}