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, = map(int, input().split())
G = [set() for _ in range(N+1)]
d = dict()
for i in range(N-1):
    a, b = map(int, input().split())
    G[a].add(b)
    G[b].add(a)
    d[(a,b)] = i
    d[(b,a)] = i
ps = []
M, = map(int, input().split())
for _ in range(M):
    a, b = map(int, input().split())
    stack = [a]
    path = []
    vs = set([a])
    while stack:
        v = stack.pop()
        if v > 0:
            path.append(v)
            if v == b:
                break
            vs.add(v)
            for u in G[v]:
                if u in vs:
                    continue
                vs.add(u)
                stack += [-v, u]
        else:
            path.pop()
    es = 0
    for i in range(len(path)-1):
        es += 1<<d[(path[i], path[i+1])]
    ps.append(es)

r=0
for i in range(2**M):
    p=m=j=0
    x=0
    while i:
        if i%2:
            x=x|ps[j]
            p+=1
        i >>= 1
        j+=1
    r+=(-1 if p%2 else 1)*(1<<(N-1-bin(x).count("1")))
print(r)
