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())
inf = 10**18
G = [[inf]*N for _ in range(N)]
edges = [[-1]*N for _ in range(N)]
for i in range(N-1):
    a, b = map(int, input().split())
    edges[a-1][b-1] = i
    edges[b-1][a-1] = i
use = [-1]*N
que = [0]
use[0] = 0
while que:
    v = que.pop()
    u = use[v]
    for nv in range(N):
        if edges[v][nv] == -1 or use[nv] != -1:
            continue
        i = edges[v][nv]
        use[nv]=u^(1<<i)
        que.append(nv)
M = int(input())
uv = [list(map(int, input().split())) for _ in range(M)]
used = [0]*(1<<M)
cnt = [0]*(1<<M)
ans = 0
for s in range(1, 1<<M):
    for i in range(M):
        if (s>>i)&1:
            u, v = uv[i]
            u, v = u-1, v-1
            p = s^(1<<i)
            used[s] = used[p]|(use[u]^use[v])
            cnt[s] = cnt[p]^1
            break
    res = 0
    for i in range(N-1):
        if used[s]&(1<<i):
            res+=1
    ans+=(-1)**cnt[s]*pow(2, N-1-res)
print(pow(2, N-1)+ans)






