Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3
# encoding:utf-8
import copy
import random
import bisect #bisect_left　これで二部探索の大小検索が行える
import fractions #最小公倍数などはこっち
import math
import sys
import collections
from decimal import Decimal # 10進数で考慮できる

mod = 10**9+7
sys.setrecursionlimit(mod) # 再帰回数上限はでdefault1000

d = collections.deque()
def LI(): return list(map(int, sys.stdin.readline().split()))

N = int(input())

graph = [[mod for i in range(N)] for i in range(N)]

weights = {}

roots = [[] for i in range(N)]
for i in range(N - 1):
    a, b, c = LI()
    a -= 1
    b -= 1
    roots[a].append(b)
    roots[b].append(a)
    weights[(a,b)] = c
    weights[(b,a)] = c

Q, K = LI()
K -= 1
xy = [0 for i in range(Q)]
for i in range(Q):
    x, y = LI()
    x -= 1
    y -= 1
    xy[i] = [x, y]
"""
xyzで全部の通りを見つけるとN ** 3で無理
なので今回は木構造に注目してBFS or DFSで解いたらいいのでは？
"""

visited = [-1 for i in range(N)]
queue = collections.deque()
queue.append(K)
visited[K] = 0

while len(queue)> 0:
    node = queue.popleft()
    cnt = visited[node]
    for n_node in roots[node]:
        if visited[n_node] == -1:
            visited[n_node] = cnt + weights[(node, n_node)]
            queue.append(n_node)
for x, y in xy:
    print(visited[x] + visited[y])


Fixed submission:
#!/usr/bin/env python3
# encoding:utf-8
import copy
import random
import bisect #bisect_left　これで二部探索の大小検索が行える
import fractions #最小公倍数などはこっち
import math
import sys
import collections
from decimal import Decimal # 10進数で考慮できる

mod = 10**9+7
sys.setrecursionlimit(mod) # 再帰回数上限はでdefault1000

d = collections.deque()
def LI(): return list(map(int, sys.stdin.readline().split()))

N = int(input())
weights = {}

roots = [[] for i in range(N)]
for i in range(N - 1):
    a, b, c = LI()
    a -= 1
    b -= 1
    roots[a].append(b)
    roots[b].append(a)
    weights[(a,b)] = c
    weights[(b,a)] = c

Q, K = LI()
K -= 1
xy = [0 for i in range(Q)]
for i in range(Q):
    x, y = LI()
    x -= 1
    y -= 1
    xy[i] = [x, y]
"""
xyzで全部の通りを見つけるとN ** 3で無理
なので今回は木構造に注目してBFS or DFSで解いたらいいのでは？
重みを単純にグラフに持つと10**5 ** 2になってしまうので注意
"""

visited = [-1 for i in range(N)]
queue = collections.deque()
queue.append(K)
visited[K] = 0

while len(queue)> 0:
    node = queue.popleft()
    cnt = visited[node]
    for n_node in roots[node]:
        if visited[n_node] == -1:
            visited[n_node] = cnt + weights[(node, n_node)]
            queue.append(n_node)

for x, y in xy:
    print(visited[x] + visited[y])

Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3
# encoding:utf-8
import copy
import random
import bisect #bisect_left　これで二部探索の大小検索が行える
import fractions #最小公倍数などはこっち
import math
import sys
import collections
from decimal import Decimal # 10進数で考慮できる

mod = 10**9+7
sys.setrecursionlimit(mod) # 再帰回数上限はでdefault1000

d = collections.deque()
def LI(): return list(map(int, sys.stdin.readline().split()))

N = int(input())

graph = [[mod for i in range(N)] for i in range(N)]
for i in range(N):
    graph[i][i] = 0

roots = [[] for i in range(N)]
for i in range(N - 1):
    a, b, c = LI()
    a -= 1
    b -= 1
    graph[a][b] = c
    graph[b][a] = c
    roots[a].append(b)
    roots[b].append(a)

Q, K = LI()
K -= 1
xy = [0 for i in range(Q)]
for i in range(Q):
    x, y = LI()
    x -= 1
    y -= 1
    xy[i] = [x, y]
"""
xyzで全部の通りを見つけるとN ** 3で無理
なので今回は木構造に注目してBFS or DFSで解いたらいいのでは？
"""

visited = [-1 for i in range(N)]
queue = collections.deque()
queue.append(K)
visited[K] = 0

while len(queue)> 0:
    node = queue.popleft()
    cnt = visited[node]
    for n_node in roots[node]:
        if visited[n_node] == -1:
            visited[n_node] = cnt + graph[node][n_node]
            queue.append(n_node)
for x, y in xy:
    print(visited[x] + visited[y])





Fixed submission:
#!/usr/bin/env python3
# encoding:utf-8
import copy
import random
import bisect #bisect_left　これで二部探索の大小検索が行える
import fractions #最小公倍数などはこっち
import math
import sys
import collections
from decimal import Decimal # 10進数で考慮できる

mod = 10**9+7
sys.setrecursionlimit(mod) # 再帰回数上限はでdefault1000

d = collections.deque()
def LI(): return list(map(int, sys.stdin.readline().split()))

N = int(input())
weights = {}

roots = [[] for i in range(N)]
for i in range(N - 1):
    a, b, c = LI()
    a -= 1
    b -= 1
    roots[a].append(b)
    roots[b].append(a)
    weights[(a,b)] = c
    weights[(b,a)] = c

Q, K = LI()
K -= 1
xy = [0 for i in range(Q)]
for i in range(Q):
    x, y = LI()
    x -= 1
    y -= 1
    xy[i] = [x, y]
"""
xyzで全部の通りを見つけるとN ** 3で無理
なので今回は木構造に注目してBFS or DFSで解いたらいいのでは？
重みを単純にグラフに持つと10**5 ** 2になってしまうので注意
"""

visited = [-1 for i in range(N)]
queue = collections.deque()
queue.append(K)
visited[K] = 0

while len(queue)> 0:
    node = queue.popleft()
    cnt = visited[node]
    for n_node in roots[node]:
        if visited[n_node] == -1:
            visited[n_node] = cnt + weights[(node, n_node)]
            queue.append(n_node)

for x, y in xy:
    print(visited[x] + visited[y])