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.

Examples

Input

5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5


Output

3
2
4


Input

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


Output

5
14
22


Input

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


Output

17000000000
# -*- coding: utf-8 -*-
#深さ優先探索　再帰
#
import sys
sys.setrecursionlimit(200000)

N=input()


a=[[] for i in range(N+1)]

#隣接リスト
for i in range(N-1):
	x,y,z=list(map(int,raw_input().split()))
	a[x].append((y,z))
	a[y].append((x,z))

#print a
	

Q,K=list(map(int,raw_input().split()))


b=[]
for i in range(Q):
	b.append(map(int, raw_input().split()))
#print b



#訪問判別フラグ
F=(N+1)*[0]

#距離
dist=[0 for i in range(N+1)]

#深さ優先探索　再帰
#現在の頂点 v, v の親 p, スタート地点Kから現在の頂点vまでの距離 d
def dfs(v,p,d):
	F[v]=1
	dist[v]=d
	for i,j in a[v]:
		#頂点 v に隣接しているかつ未訪問 
		if F[i]==0:
			#DFS(次の頂点i, 次に親となる頂点v, d+ 頂点vと頂点iの間の辺のコスト)
			dfs(i,v,d+j)


dfs(K,-1,0)

for x in range(len(b)):
	print dist[ b[x][0] ]+ dist[ b[x][1] ]

