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
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from collections import deque,defaultdict

def all_bit_count(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
  
n = int(readline())
d = defaultdict(int)
cnt = 1

g = [[] * n for i in range(n)]
for i in range(n-1):
  u, v = map(int,readline().split())
  u -= 1
  v -= 1
  g[u].append(v)
  g[v].append(u)

m = int(readline())
c = []
for i in range(m):
  u, v = map(int,readline().split())
  u -= 1
  v -= 1  
  que = deque()
  que.append((u,0))
  chk = [False]*n
  chk[u] = True
  while que:
      now,L = que.popleft()
      if now == v:
        c.append(L)
        break
      for i in g[now]:
          if not chk[i]:
            chk[i] = True
            mi = min(now,i)
            ma = max(now,i)
            mm = mi*100+ma
            if d[mm] == 0:
              d[mm] = cnt
              cnt += 1
            que.append((i,L|(1<<d[mm])))
                     
import itertools
ans = 1<<(n-1)
for i in range(1,m+1):
  for j in itertools.combinations(c,i):
    K = 0
    for k in j:
      K |= k
    l = all_bit_count(K)
    ans += pow(-1,i)*pow(2,n-1-l)
    
print(ans)