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
N = int(input())
ab = [list(map(int, input().split())) for _ in range(N - 1)]
M = int(input())
uv = [list(map(int, input().split())) for _ in range(M)]

class Tree:
  C, RL = {}, {}
  R, N, D, S, P = None, None, None, None, None
  def __init__(s, num):
    s.N = num
  def set(s, a, b):
    if a in s.C: s.C[a].append(b)
    else: s.C[a] = [b]
    if b in s.C: s.C[b].append(a)
    else: s.C[b] = [a]
  def dfs(s, x, y, r):
    if x == y:
      return [x]
    for i in s.C[x]:
      if i != r:
        t = s.dfs(i, y, x)
        if t != False:
          return [x] + t
    return False

T = Tree(N)
for a, b in ab:
  a -= 1
  b -= 1
  T.set(a, b)

D = [[-1] * N for _ in range(N)]
for i in range(N - 1):
  a, b = ab[i][0] - 1, ab[i][1] - 1
  a, b = min(a, b), max(a, b)
  D[a][b] = i

NL = [[] for _ in range(M)]
for i in range(M):
  x, y, r = uv[i][0] - 1, uv[i][1] - 1, uv[i][0] - 1
  L = T.dfs(x, y, r)
  n = L[0]
  t = 0
  for j in range(1, len(L)):
    m = L[j]
    x, y = min(n, m), max(n, m)
    t = t | (1 << D[x][y])
    n = m
  NL[i] = t
  

ans = 1 << (N - 1)
c = 0
for i in range(1, 1 << M):
  t = i
  f = 0
  k = 0
  for j in range(M):
    if t & 1:
      f += 1
      k = k | NL[j]
    t >>= 1
  cnt = bin(k).count("1")
  if f % 2 == 1:
    c += pow(2, (N - 1 - cnt))
  else:
    c -= pow(2, (N - 1 - cnt))
print(ans - c)