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
def main():
  import sys
  input = sys.stdin.readline
  
  N = int(input())
  E = [[] for _ in range(N)]
  # E:頂点a,bをつなぐ枝にiという番号をふる。
  for i in range(N-1):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    E[a].append((b,i))
    E[b].append((a,i))
  
  # dfs(u,v,p):uからvまでのパス上にある頂点をEGにセットする。
  EG = []
  def dfs(u,v,p):
    if u == v:
      return True
    for t,i in E[u]:
      if t == p:
        continue
      if dfs(t,v,u):
        EG.append(i)
        return True
    return False
  
  M = int(input())
  EM = []
  for i in range(M):
    u,v = map(int,input().split())
    u -= 1
    v -= 1
    EM.append((u,v))

  # ES[i]:i番目の制約のパス上にある頂点
  ES = []
  for i in range(M):
    u,v = EM[i]
    EG = []
    dfs(u,v,-1)
    ES.append((u,v,EG))
  
  ans = 0
  for i in range(1<<M):
    # EG2:iのビットが立っているところの制約のパス上にある頂点の集合。
    # bc:iのビットが立っている個数。
    EG2 = []
    bc = 0
    for j in range(M):
      if (i>>j)&1 == 1:
        bc += 1
        EG2 += ES[j][2]
    p = len(set(EG2))
    # 包除原理より
    if bc%2 == 0:
      ans += 2**(N-1-p)
    else:
      ans -= 2**(N-1-p)
  print(ans)

if __name__ == '__main__':
  main()
