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
import sys
input = sys.stdin.readline

N = int(input())
G = [[] for _ in range(N+1)]
for i in range(N-1):
    a, b = map(int, input().split())
    G[a].append((b, i))
    G[b].append((a, i))


def get_path(u, v):
    stack = [u]
    prev = [None] * (N+1)
    visited = [False] * (N+1)
    visited[u] = True
    while stack:
        node = stack.pop()
        for next, i in G[node]:
            if not visited[next]:
                visited[next] = True
                prev[next] = (node, i)
                stack.append(next)
    path = 0
    node = v
    while node != u:
        p, i = prev[node]
        node = p
        path += 1 << i
    return path


def count(bits, N):
    cnt = 0
    for i in range(N):
        if ((bits >> i) & 1) == 1:
            cnt += 1
    return cnt


M = int(input())
uv = [tuple(map(int, input().split())) for _ in range(M)]
paths = [get_path(u, v) for u, v in uv]
ans = 2**(N-1)
for bit in range(1, 1 << M):
    cnt = 0
    tmp = 0
    for i in range(M):
        if ((bit >> i) & 1) == 1:
            cnt += 1
            tmp |= paths[i]
    ans += (-1)**cnt * 2**(N-1-count(tmp, N-1))
print(ans)
