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
sys.setrecursionlimit(10**8)
input = sys.stdin.readline

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


def dfs(v, v_p, target):
    if v == target:
        return True, 0

    path = 0
    for v_next, edge_idx in graph[v]:
        if v_next == v_p:
            continue
        is_target, path = dfs(v_next, v, target)
        if is_target:
            return is_target, path | (1 << edge_idx)
    return False, 0


m = int(input())
cond = []
for _ in range(m):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    _, path = dfs(u, -1, v)
    # print(u, v, bin(path))
    cond.append((u, v, path))

ans = 1 << (n - 1)
for bit in range(1, 1 << m):
    cnt = [0] * n
    l = 0
    path = 0
    for i, (u, v, path_i) in enumerate(cond):
        if (bit >> i) & 1:
            l += 1
            path = path | path_i

    # print(bin(bit), cnt)
    c = n - 1 - bin(path).count('1')
    # print(bin(bit), cnt, c, l)
    ans += ((-1)**(l % 2)) * (1 << c)

print(ans)
