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)]
edge = {}


for i in range(n-1):
    a, b = map(int, input().split())
    g[a-1].append(b-1)
    g[b-1].append(a-1)
    edge[(a-1, b-1)] = i
    edge[(b-1, a-1)] = i

m = int(input())


def dfs(u,v):
    visited = [False]*n
    visited[u] = True
    stack = [u]
    prev = [-1]*n
    while stack:
        uu = stack.pop()
        for vv in g[uu]:
            if not visited[vv]:
                visited[vv] = True
                prev[vv] = uu
                stack.append(vv)
        if prev[v] != -1:
            break
    bin_edge = 0
    head = v
    while head != u:
        bin_edge += 1 << edge[(head, prev[head])]
        head = prev[head]
    return bin_edge

intv = [0]*m

for i in range(m):
    u, v = map(int, input().split())
    intv[i] = dfs(u-1,v-1)


mm = 1 <<m

ans = 0

for i in range(mm):
    a = 0
    cnt = 0
    for j in range(m):
        if (i >> j) & 1:
            cnt += 1
            a = a | intv[j]
    white = str(bin(a)).count('1')
    ans += ((-1)**(cnt))*2**(n-1-white)

print(abs(ans))
