from collections import deque
N = int(input())
E = [[] for _ in range(N)]
for _ in range(N-1):
    a,b,c = map(int, input().split())
    E[a-1].append(c*N+b-1)
    E[b-1].append(c*N+a-1)


par = [-1]*N
rnk = [-1]*N
par[0] = 0
rnk[0] = 0
dist = [0]*N
q = deque()
q.append((0, 0, 0))
while q:
    temp = q.popleft()
    v = temp[0]
    r = temp[1]
    d = temp[2]
    for j in E[v]:
        if par[j%N] < 0:
            par[j%N] = v
            rnk[j%N] = r+1
            dist[j%N] = d+j//N
            q.append((j%N, r+1, d+j//N))

#2^k上の祖先を求める
par[0] = 0
LV = (N-1).bit_length()
kpar = [par]
for k in range(LV):
    T = [0]*N
    for i in range(N):
        if par[i] == 0:
            continue
        T[i] = par[par[i]]
    kpar.append(T)
    par = T

#LCAをO(logN)で求める
def lca(u, v):
    if rnk[u] > rnk[v]:
        u, v = v, u
    d = rnk[v]-rnk[u]
    for i in range(LV+1):
        if d & 1:
            v = kpar[i][v]
        d >>= 1
    if u == v:
        return u
    for k in range(LV-1, -1, -1):
        pu = kpar[k][u]
        pv = kpar[k][v]
        if pu != pv:
            u = pu
            v = pv
    return kpar[0][u]


Q = int(input())
for _ in range(Q):
    s,t = map(int, input().split())
    s -= 1
    t -= 1
    an = lca(s, t)
    print(dist[s]+dist[t]-2*dist[an])