We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?

* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.

Constraints

* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M


Output

Print the number of ways to paint the edges that satisfy all of the M conditions.

Examples

Input

3
1 2
2 3
1
1 3


Output

3


Input

2
1 2
1
1 2


Output

1


Input

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


Output

9


Input

8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8


Output

62
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
from collections import deque
from fractions import gcd
from functools import lru_cache


#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')

#############
# Functions #
#############

######INPUT######
def inputI(): return int(input().strip())
def inputS(): return input().strip()
def inputIL(): return list(map(int,input().split()))
def inputSL(): return list(map(str,input().split()))
def inputILs(n): return list(int(input()) for _ in range(n))
def inputSLs(n): return list(input().strip() for _ in range(n))
def inputILL(n): return [list(map(int, input().split())) for _ in range(n)]
def inputSLL(n): return [list(map(str, input().split())) for _ in range(n)]

#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n):
    return kaijo_memo[n]
  if(len(kaijo_memo) == 0):
    kaijo_memo.append(1)
  while(len(kaijo_memo) <= n):
    kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n):
    return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0):
    gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n):
    gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if(n == r):
    return 1
  if(n < r or r < 0):
    return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])
    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])
    return arr

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#############
# Main Code #
#############

N = inputI()

graph = [[] for _ in range(N)]
edges = []
 
for _ in range(N - 1):
  a, b = inputIL()
  edges.append((a-1, b-1))
  
for a, b in edges:
  graph[a].append(b)
  graph[b].append(a)
  
  
M = inputI()
m = inputILL(M)

prv = [None for i in range(N)]

prv[0] = -1

q = deque([0])

while q:
  u = q.popleft()
  for v in graph[u]:
    if prv[v] == None:
      prv[v] = u
      q.append(v)

to_root = []

for i in range(N):
  s = 0
  now = i
  while now!=0:
    s += pow(2,now)
    now = prv[now]
  to_root.append(s)

  
ans = 0
for k in range(pow(2,M)):
  J = []
  for i in range(M):
    if k>>i & 1:
      J.append(m[i])
  mask = 0
  for u,v in J:
    mask |= to_root[u-1]^to_root[v-1]
  ##########
  count = 0
  while mask:
    mask &= mask -1
    count += 1
  ##########
  c = 0
  while k:
    k &= k -1
    c += 1
  sign = 1
  if c%2 == 1:
    sign = -1
  ###########
  ans += sign * pow(2,N-1-count)

print(ans)