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=[list(map(int,input().split())) for i in range(N-1)]
M=int(input())
uv=[list(map(int,input().split())) for i in range(M)]
data=[[] for i in range(N+1)]
for a,b in ab:
    data[a].append(b)
    data[b].append(a)
node=[[-1]*(N+1) for i in range(N-1)]
for i in range(N-1):
    a,b=ab[i]
    flag=[0]*(N+1)
    flag[a]=1
    flag[b]=1
    node[i][a]=0
    que=[a]
    while que:
        u=que.pop()
        for v in data[u]:
            if flag[v]==0:
                flag[v]=1
                node[i][v]=0
                que.append(v)
    node[i][b]=1
    que=[b]
    while que:
        u=que.pop()
        for v in data[u]:
            if flag[v]==0:
                flag[v]=1
                node[i][v]=1
                que.append(v)
dp=[[0]*2**M for i in range(N)]
dp[0][0]=1
for i in range(N-1):
    h=node[i]
    num=0
    for j in range(M):
        u,v=uv[j]
        if h[u]==h[v]:
            continue
        else:
            num+=2**j
    dp[i+1]=dp[i][:]
    for j in range(2**M):
        dp[i+1][j|num]+=dp[i][j]
print(dp[-1][-1])