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
from collections import deque
n = int(input())
Graph = [[] for _ in range(n)]
edge = {}
for i in range(n-1):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    Graph[a].append(b)
    Graph[b].append(a)
    edge[1000*a+b] = i
    edge[1000*b+a] = i

data = []

def f(u,v):
    prev = [u]*n
    used = [False]*n
    que = deque([u])
    while que:
        now = que.popleft()
        for nex_v in Graph[now]:
            if used[nex_v]:continue
            prev[nex_v] = now
            used[nex_v] = True
            que.append(nex_v)
            if nex_v == v:
                break
    path = [v]
    path_edge = []
    t = v
    while prev[t] != u:
        path.append(prev[t])
        path_edge.append(edge[1000*t+prev[t]])
        t = prev[t]
    path_edge.append(edge[1000*t+prev[t]])
    path.append(u)
    return path_edge

m = int(input())
for i in range(m):
    u,v = map(int,input().split())
    u -= 1
    v -= 1
    path = f(u,v)
    data.append(path[::-1])

#print(data)
ans = 2**(n-1)
for i in range(1,1<<m):
    now = []
    for j in range(m):
        if i>>j&1:
            now.append(j)

    this = [True]*(n-1)
    for j in now:
        for k in data[j]:
            this[k] = False

    ponta = 1
    q = 0
    for j in range(n-1):
        if this[j]:
            ponta *= 2
            q += 1

    if len(now)%2 == 0:
        ans += ponta
    else:
        ans -= ponta

print(ans)