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 _ in [0]*(N-1)]
M = int(input())
UV = [list(map(int,input().split())) for _ in [0]*M]

E = [[] for _ in [0]*N]
for a,b in AB:
    a-=1
    b-=1
    E[a].append(b)
    E[b].append(a)

P = [-1]*N
P[0] = 0
q = [0]
while q:
    i = q.pop()
    for j in E[i]:
        if P[j]!=-1: continue
        P[j] = i
        q.append(j)
P[0] = -1

def dist_bfs(N,E,start):
    d = [-1]*N
    d[start] = 0
    q = [start]
    while q:
        qq = []
        for i in q:
            di = d[i]
            for j in E[i]:
                if d[j]!=-1:continue
                d[j] = di+1
                q.append(j)
        q = qq
    return d

path = []
for u,v in UV:
    u -= 1
    v -= 1
    tmp = []
    d = dist_bfs(N,E,u)
    q = [v]
    while q:
        i = q.pop()
        di = d[i]
        for j in E[i]:
            if d[j] == di-1:
                q.append(j)
                if P[i]==j:
                    tmp.append(i)
                else:
                    tmp.append(j)
    path.append(set(tmp))

ans = 2**(N-1)
for i in range(1,2**M):
    tmp = set(range(1,N))
    cnt = 0
    for j in range(M):
        if (i>>j)&1==0:continue
        cnt += 1
        tmp -= path[j]
    ans += ((-1)**cnt) * (2**(len(tmp)))
print(ans)