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 itertools import combinations

n = int(input())
ab = [list(map(int, input().split())) for _ in range(n - 1)]
m = int(input())
uv = [list(map(int, input().split())) for _ in range(m)]

adj = [[] for _ in range(n)]
ids = [[-1] * n for _ in range(n)]
for i, (a, b) in enumerate(ab):
    a -= 1
    b -= 1
    adj[a].append(b)
    adj[b].append(a)
    ids[a][b] = i
    ids[b][a] = i


def dfs(s, g):
    d = [-1] * n
    d[s] = 0
    p = [-1] * n
    stack = [s]

    while stack:
        u = stack.pop()
        for v in adj[u]:
            if d[v] == -1:
                d[v] = d[u] + 1
                p[v] = u
                stack.append(v)

    v = g
    ret = 0
    while v != s:
        i = ids[v][p[v]]
        ret += 1 << i
        v = p[v]

    return ret


def bin_count(tpl):
    sm_or = 0
    for e in tpl:
        sm_or |= e

    return bin(sm_or).count("1")


path = [dfs(u - 1, v - 1) for u, v in uv]
ex = 0
for i in range(1, m + 1):
    for tpl in combinations(path, i):
        ex += pow(2, n - 1 - bin_count(tpl)) * (-1) ** (i % 2 + 1)

ans = pow(2, n - 1) - ex
print(ans)
