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())
X = [[] for i in range(N)]
for i in range(N-1):
    x, y = map(int, input().split())
    X[x-1].append(y-1)
    X[y-1].append(x-1)

P = [-1] * N
DE = [0] * N
Q = [0]
while Q:
    i = Q.pop()
    for a in X[i][::-1]:
        if a != P[i]:
            P[a] = i
            DE[a] = DE[i] + 1
            X[a].remove(i)
            Q.append(a)

def lp(u, v):
    t = 0
    while u != v:
        if DE[u] > DE[v]:
            t += 1 << u-1
            u = P[u]
        elif DE[u] < DE[v]:
            t += 1 << v-1
            v = P[v]
        else:
            t += 1 << u-1
            t += 1 << v-1
            u = P[u]
            v = P[v]
    
    return t

Y = []
M = int(input())
for _ in range(M):
    a, b = map(int, input().split())
    a, b = a-1, b-1
    Y.append(lp(a, b))

D = {1<<i: i for i in range(50)}
Z = [0] * (1<<M)
ans = 0
CC = [0] * N
BC = [0] * (1<<M)
for m in range(1, 1<<M):
    a = m & (-m)
    BC[m] = BC[m^a] + 1
for m in range(1<<M):
    a = m & (-m)
    if a == m:
        if a == 0:
            Z[m] = 0
        else:
            Z[m] = Y[D[a]]
    else:
        Z[m] = Z[m^a] | Y[D[a]]
    
    CC[N - 1 - bin(Z[m]).count("1")] += (1 if BC[m] % 2 == 0 else -1)

print(sum([2 ** i * CC[i] for i in range(N)]))