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

def main():
  n = int(input())
  T = [[] for _ in range(n)]
  D = {}
  for i in range(n-1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    T[a].append(b)
    T[b].append(a)
    D[(a, b)] = 1<<i
    D[(b, a)] = 1<<i
  m = int(input())

  def dfs(u, v):
    parent = [-1]*n
    seen = [False]*n
    stack = [u]
    seen[u] = True
    while stack:
      nv = stack[-1]
      if nv == v:
        return stack
      for i in T[nv]:
        if seen[i]:
          continue
        stack.append(i)
        seen[i] = True
        break
      else:
        stack.pop()

  def popcount(n):
    c = (n & 0x5555555555555555) + ((n>>1) & 0x5555555555555555)
    c = (c & 0x3333333333333333) + ((c>>2) & 0x3333333333333333)
    c = (c & 0x0f0f0f0f0f0f0f0f) + ((c>>4) & 0x0f0f0f0f0f0f0f0f)
    c = (c & 0x00ff00ff00ff00ff) + ((c>>8) & 0x00ff00ff00ff00ff)
    c = (c & 0x0000ffff0000ffff) + ((c>>16) & 0x0000ffff0000ffff)
    c = (c & 0x00000000ffffffff) + ((c>>32) & 0x00000000ffffffff)
    return c
              
  E = [0]*m
  for i in range(m):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    edge = dfs(u, v)
    for j in range(len(edge)-1):
      E[i] |= D[(edge[j], edge[j+1])]

  ans = 0
  for i in range(2**m):
    s = 0
    for j in range(m):
      if i>>j & 1:
        s |= E[j]
    cnt = popcount(i) % 2
    ans += (-1)**cnt * 2**(n-1-popcount(s))
  print(ans)
  
if __name__ == "__main__":
  main()