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())
ab = [tuple(map(int,input().split())) for _ in range(N - 1)]
M = int(input())
uv = [tuple(map(int,input().split())) for _ in range(M)]
route = [[] for _ in range(N)]

for i, (a, b) in enumerate(ab):
    a -= 1
    b -= 1
    route[a].append((b, i))
    route[b].append((a, i))

path = [0] * M
for i, (u, v) in enumerate(uv):
    u -= 1
    v -= 1
    stack = [u]
    visited = [0] * N
    visited[u] = 1
    dire = [[] for _ in range(N)]
    while True:
        s = stack.pop()
        for g, j in route[s]:
            if visited[g]:
                continue   
            stack.append(g)
            dire[g].append((s, j))
            visited[g] = 1
        if visited[v]:
            break
    s = v
    while True:
        g, num = dire[s][0]
        path[i] |= (1 << num)
        if g == u:
            break
        s = g

data = [0] * (1 << M)
ans = 2 ** (N - 1)
for i in range(1, 1 << M):
    k = len(bin(i)) - 3
    data[i] = data[i - (2 ** k)] | path[k]
    power = bin(data[i]).count("1")
    judge = bin(i).count("1")
    ans += (-2 ** (N - 1 - power) if judge % 2 else 2 ** (N - 1 - power))
print(ans)
