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
INF = 1 << 30

n = int(input())

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

m = int(input())

u = [0 for i in range(m)]
v = [0 for i in range(m)]
ls = [0 for i in range(m)]
for i in range(m):
    u[i], v[i] = map(int, input().split())
    u[i] -= 1
    v[i] -= 1

    cost = [INF for j in range(n)]
    cost[u[i]] = 0
    st = [u[i]]
    pre = [-1 for j in range(n)]
    pre[u[i]] = -2

    while len(st) > 0:
        now = st.pop()
        for to in g[now]:
            if pre[to] == -1:
                pre[to] = now
                st.append(to)
    
    pos = v[i]
    while pos != u[i]:
        for j in range(len(g[pos])):
            to = g[pos][j]
            if to == pre[pos]:
                ls[i] |= 1<<num[pos][j]
                pos = to
                break

ans = 0

for bit in range(1<<m):
    cnt = 0
    val = 0
    for i in range(m):
        if ((bit>>i)&1) == 1:
            cnt += 1
            val |= ls[i]
    
    pcnt = 0
    for i in range(n):
        if ((val>>i)&1) == 1:
            pcnt += 1
    
    nokori = n-1-pcnt

    if cnt%2 == 0:
        ans += 2**nokori+1
    else:
        ans -= 2**nokori+1

print(ans)