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
import sys
input=sys.stdin.readline
n=int(input())
Edge=[[] for i in range(n)]
for i in range(n-1):
  a,b=map(int,input().split())
  Edge[a-1].append((b-1,i))
  Edge[b-1].append((a-1,i))
m=int(input())
M=[]
for i in range(m):
  M.append(tuple(map(int,input().split())))

inf=float("inf")
path=[-1]*n #根から各点までのパスが通過する辺をbitで管理したもの
path[0]=0
stack=[0]
while stack:
  cur=stack.pop()
  for next,num in Edge[cur]:
    if path[next]==-1:
      path[next]=path[cur]|(1<<num)
      stack.append(next)

def popcount(x):
  return bin(x).count("1")

passing=[] #Mの各u,vについて、通過する辺をbitで持ったもの
for u,v in M:
  passing.append((path[u-1]^path[v-1]))

ans=2**(n-1)
#Mのすべての部分集合について符号付で足し合わせる
for i in range(1,2**m):
  i_ct=popcount(i)#満たすべき条件の数、これで符号を決める
  condition=0
  for j in range(m):
    if i&(1<<j):#j番目の条件を満たす
      condition=condition|passing[j]
  cond_ct=popcount(condition) #iが表すMの部分集合の条件をすべて満たす場合にすべて白になる辺集合の要素数

  ans+=pow((-1),i_ct)*pow(2,n-1-cond_ct)
print(ans)