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:
def dfs(s, t, links):
    visited = set()
    q = [(s, 0)]
    while q:
        v, used = q.pop()
        if v == t:
            return used
        visited.add(v)
        for lb, u in links[v]:
            if u in visited:
                continue
            q.append((u, used | lb))


n = int(input())
links = [set() for _ in range(n)]
for i in range(n - 1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    lb = 1 << i
    links[a].add((lb, b))
    links[b].add((lb, a))
conditions = []
m = int(input())
for i in range(m):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    conditions.append(dfs(u, v, links))

ans = 0
for i in range(1 << m):
    cnd_cnt = bin(i).count('1')
    bit = 0
    k = i
    while k:
        b = k & -k
        j = b.bit_length() - 1
        bit |= conditions[j]
        k ^= b
    free_link_cnt = n - 1 - bin(bit).count('1')
    val = 2 ** free_link_cnt
    if cnd_cnt % 2:
        ans -= val
    else:
        ans += val
print(ans)