Problem: 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

Solution (in PYTHON3):

Solution:
from collections import defaultdict, Counter

N = int(input())
ab = [[] for _ in range(N)]
for i in range(N - 1) :
  a, b = map(int, input().split())
  ab[a-1].append((b-1, i))
  ab[b-1].append((a-1, i))

M = int(input())
uv = [[int(i) for i in input().split()] for _ in range(M)]

cnt = [set() for _ in range(M)]
def dfs(cur, pre, dest, i) :
  if cur == dest :
    return True
  
  for nex, j in ab[cur] :
    if nex == pre :
      continue
    if dfs(nex, cur, dest, i) :
      cnt[i].add(j)
      return True

  return False

for i, (u, v) in enumerate(uv) :
  dfs(u - 1, -1, v - 1, i)

ret = 1 << (N - 1)
for i in range(1, 1 << M) :
  sgn = 1
  s = set([])
  for j in range(M) :
    if i & (1 << j) :
      s |= cnt[j]
      sgn *= -1
  ret += sgn * (1 << (N - 1 - len(s)))

print(ret)