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 collections import defaultdict


def get_path(from_node, to_node, adj, nodes2edge):
    # from_nodeからto_nodeまでの最短パスで通るedgeを返す

    def dfs(node, parent, edges):
        if node == to_node:
            return edges
        min_path = None
        for child in adj[node]:
            if child == parent:
                continue
            path = dfs(child, node, edges + [nodes2edge[(node, child)]])
            if path is None:
                continue
            if min_path is None or len(min_path) > len(path):
                min_path = path
        return min_path

    return dfs(from_node, None, list())


def count_bits(n):
    return len([i for i in range(len('{:b}'.format(n))) if (n >> i) & 1 == 1])


def main():
    N = int(input())
    adj = defaultdict(list)
    nodes2edge = defaultdict(lambda :-1)
    for i in range(N - 1):
        a, b = list(map(lambda x: int(x) - 1, input().split(' ')))
        adj[a].append(b)
        adj[b].append(a)
        nodes2edge[(a, b)] = i
        nodes2edge[(b, a)] = i
    M = int(input())
    constraints = [list(map(lambda x: int(x) - 1, input().split(' '))) for _ in range(M)]
    # 各制約を「パス u->v 上の辺はすべて白」として扱い、包除原理で計算
    base = [0] * M  # 各制約で出てくる最短 u->v パス で通るedgeの2進表記
    for m, constraint in enumerate(constraints):
        u, v = constraint
        path = get_path(u, v, adj, nodes2edge)
        v = sum([2**e for e in path])
        base[m] = v
    dp = [0] * (2**M)  # 制約の集合で出てくる最短 u->v パスで通るedgeの和集合の2進表記
    ans = 2**(N - 1)
    for S in range(1, 2**M):
        lenbit_s = len('{:b}'.format(S))
        dp[S] = dp[S - (1 << (lenbit_s - 1))] | base[lenbit_s - 1]
        cnt_s = count_bits(S)
        cnt_dp = count_bits(dp[S])
        ans += (-1)**(cnt_s % 2) * 2**(N - 1 - cnt_dp)
    print(ans)


if __name__ == '__main__':
    main()
