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())
peer = [[] for _ in range(n)]
for _ in range(n-1):
  a,b = map(int,input().split())
  a -= 1
  b -= 1
  peer[a].append(b)
  peer[b].append(a)
edge = [0 for _ in range(n)]
seen = [0 for _ in range(n)]
seen[0] = 1
now = [0]
while now:
  last = now
  now = []
  for x in last:
    for y in peer[x]:
      if seen[y] == 0:
        edge[y] = edge[x] + 2**(y-1)
        seen[y] = 1
        now.append(y)

def count1(x):
  y = bin(x)
  z = 0
  l = len(y)
  for li in range(l):
    if y[li] == '1':
      z += 1
  return z

def count2(x):
  z = 0
  l = len(x)
  for li in range(l):
    if x[li] == '1':
      z += 1
  return z


m = int(input())
restr = []
for _ in range(m):
  u,v = map(int,input().split())
  u -= 1
  v -= 1
  restr.append(edge[u]^edge[v])
#print(edge)
#print(restr)

ans = 0
for i in range(2**m):
  b = bin(i)[2:]
  time = '0' * (m-len(b)) + b
  tmp = 0
  for j in range(m):
    if time[j] == '1':
      tmp |= restr[j]
  if count2(time) % 2 == 0:
    ans += 2**(n-1-count1(tmp))
    #print('+',count2(time),time,count1(tmp),ans,bin(tmp))
  else:
    ans -= 2**(n-1-count1(tmp))
    #print('-',count2(time),time,count1(tmp),ans,bin(tmp))
#print(restr)
print(ans)