### BUGGY CODE ###

import bisect
import cmath
import heapq
import itertools
import math
import operator
import os
import random
import re
import string
import sys
from collections import Counter, deque, defaultdict
from copy import deepcopy
from decimal import Decimal
from fractions import gcd
from functools import lru_cache, reduce
from operator import itemgetter, mul, add, xor

import numpy as np

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7


# MOD = 998244353


class DoublingLCA:
    """
    LCA ダブリング版
    初期化 O(NlogN)、クエリ O(logN)
    """

    def __init__(self, graph, max_v, root):
        """
        :param list of (list of int) graph:
        :param int max_v:
        :param int root:
        """
        self.graph = graph
        self.size = max_v + 1
        self.root = root

        # AtCoder の PyPy 2.4.0 では math.log2 が使えない
        self.MAX_LOG_V = math.floor(math.log(self.size, 2)) + 1
        # depths[v]: v の root からの距離
        self.depths = [-1] * self.size
        # parents[k][v]: 親に 2^k たどった頂点
        self.parents = [[-1] * self.size for _ in range(self.MAX_LOG_V)]

        self._init()

    def _init(self):
        # depths と parents[0] を初期化
        seen = [False] * self.size
        stack = [(self.root, 0, -1)]
        while stack:
            v, d, par = stack.pop()
            self.parents[0][v] = par
            self.depths[v] = d
            seen[v] = True
            for u in self.graph[v]:
                if not seen[u]:
                    stack.append((u, d + 1, v))

        # 各 parents を初期化
        for k in range(self.MAX_LOG_V - 1):
            for v in range(self.size):
                if self.parents[k][v] < 0:
                    # 親がなければ -1
                    self.parents[k + 1][v] = -1
                else:
                    self.parents[k + 1][v] = self.parents[k][self.parents[k][v]]

    def lca(self, u, v):
        """
        :param int u:
        :param int v:
        """
        # 深さを合わせる
        if self.depths[u] > self.depths[v]:
            u, v = v, u
        for k in range(self.MAX_LOG_V):
            if (self.depths[v] - self.depths[u]) >> k & 1:
                v = self.parents[k][v]
        if v == u:
            return v

        # にぶたん
        for k in reversed(range(self.MAX_LOG_V)):
            if self.parents[k][u] != self.parents[k][v]:
                u = self.parents[k][u]
                v = self.parents[k][v]
        return self.parents[0][u]

    def distance(self, u, v):
        """
        u, v 間の距離
        depth[u] + depth[v] - depth[lca] * 2
        :param u:
        :param v:
        :rtype: int
        """
        lca = self.lca(u, v)
        return self.depths[u] + self.depths[v] - self.depths[lca] * 2


# 包除原理つかう

N = int(sys.stdin.buffer.readline())
AB = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N - 1)]
M = int(sys.stdin.buffer.readline())
VU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(M)]

graph = [[] for _ in range(N)]
edges_i = {}
for i, (a, b) in enumerate(AB):
    a -= 1
    b -= 1
    graph[a].append(b)
    graph[b].append(a)
    edges_i[a, b] = i
    edges_i[b, a] = i

root = 0
parents = [None] * N
stack = [(root, None)]
while stack:
    v, p = stack.pop()
    parents[v] = p
    for u in graph[v]:
        if u == p:
            continue
        stack.append((u, v))

lca = DoublingLCA(graph=graph, max_v=N - 1, root=root)
constraints = [0] * M
# 制約ごとにどの辺を使うかを持つ
for i, (v, u) in enumerate(VU):
    v -= 1
    u -= 1
    ancestor = lca.lca(v, u)
    while v != ancestor:
        constraints[i] |= 1 << edges_i[v, parents[v]]
        v = parents[v]
    while u != ancestor:
        constraints[i] |= 1 << edges_i[u, parents[u]]
        u = parents[u]


# for c in constraints:
#     print(np.binary_repr(c, 10))


def count(edges_bin):
    cnt = 0
    while edges_bin > 0:
        cnt += edges_bin & 1
        edges_bin >>= 1
    return pow(2, N - 1 - cnt)


# 余事象を数える
# 制約をいくつか選んだときに、1つ以上の制約を満たさない場合の数
# 1つの制約を満たさない数は、2^(制約と関係ない辺の数) == 2^((N - 1) - 使う辺の数)
ans = 0
for choice in range(1, 1 << M):
    cnt = 0
    edges = 0
    for i in range(choice.bit_length()):
        if choice >> i & 1:
            cnt += 1
            edges |= constraints[i]
    ans += count(edges) * (-1) ** (cnt - 1)
print(2 ** (N - 1) - ans)


### FIXED CODE ###

import os
import sys
from functools import lru_cache

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353

N = int(sys.stdin.buffer.readline())
AB = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N - 1)]
M = int(sys.stdin.buffer.readline())
VU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(M)]

# N <= 50
# M <= 20


graph = [[] for _ in range(N)]
edges = {}
for i, (a, b) in enumerate(AB):
    a -= 1
    b -= 1
    graph[a].append(b)
    graph[b].append(a)
    edges[a, b] = i
    edges[b, a] = i


@lru_cache(maxsize=None)
def path(v, u, p):
    if v == u:
        return [u]
    for nv in graph[u]:
        if nv == p:
            continue
        ret = path(v, nv, u)
        if ret is not None:
            return ret + [u]
    return None


# 辺ごとに、黒くしたときにどの制約を満たすことができるかを持つ
constraints = [0 for _ in range(N - 1)]
for i, (v, u) in enumerate(VU):
    v -= 1
    u -= 1
    pp = path(v, u, None)
    for pv, pu in zip(pp, pp[1:]):
        constraints[edges[pv, pu]] |= 1 << i

# for c in constraints:
#     import numpy as np
#
#     print(np.binary_repr(c, M))

# or したときに全部 True になる組み合わせの数
dp = [0 for _ in range(1 << M)]
dp[0] = 1
for c in constraints:
    for i in reversed(range(1 << M)):
        dp[i | c] += dp[i]
print(dp[-1])