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
n = int(input())
g = [[] for _ in range(n)]
e = []
for _ in range(n - 1):
    a, b = map(int, input().split())
    a, b = min(a, b), max(a, b)
    g[a - 1].append(b - 1)
    g[b - 1].append(a - 1)
    e.append((a - 1, b - 1))
m = int(input())
l = [set() for _ in range(m)]
for i in range(m):
    u, v = map(int, input().split())
    s = [u - 1]
    dist = [-1] * n
    dist[u - 1] = 0
    r = [-1] * n
    while s:
        d = s.pop()
        for node in g[d]:
            if dist[node] == -1:
                s.append(node)
                dist[node] = dist[d] + 1
                r[node] = d
    d = v - 1
    while r[d] != -1:
        a, b = min(d, r[d]), max(d, r[d])
        l[i].add(e.index((a, b)))
        d = r[d]
ans = 0
for i in range(2 ** m):
    cnt = 0
    s = set()
    for j in range(m):
        if (i >> j) & 1:
            cnt += 1
            s |= l[j]
    if cnt % 2:
        ans -= 2 ** (n - len(s) - 1)
    else:
        ans += 2 ** (n - len(s) - 1)
print(ans)
