Problem: 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

Solution (in PYTHON3):

Solution:
# https://drken1215.hatenablog.com/entry/2020/01/26/164200
# bit_pathの参考にした

def main():
    from collections import namedtuple
    from functools import reduce
    from itertools import combinations
    from operator import or_

    Edge = namedtuple('Edge', 'to edge_id')

    def bit_path(curr, goal, par=-1):
        if curr == goal:
            return 0
        for e in g[curr]:
            if e.to == par: continue
            res = bit_path(curr=e.to, goal=goal, par=curr)
            if ~res:  # -1以外
                return res | (1 << e.edge_id)
        return -1

    N = int(input())

    g = tuple(set() for _ in range(N))
    for edge_id in range(N - 1):
        a, b = (int(x) - 1 for x in input().split())
        g[a].add(Edge(to=b, edge_id=edge_id))
        g[b].add(Edge(to=a, edge_id=edge_id))

    M = int(input())

    cons = []
    for _ in range(M):
        u, v = (int(x) - 1 for x in input().split())
        b = bit_path(u, v)
        cons.append(b)

    ans = 1 << (N - 1)
    for k in range(1, M + 1):
        for comb in combinations(cons, r=k):
            b = reduce(or_, comb)
            ans += ((-1) ** k) * (1 << (N - 1 - bin(b).count('1')))
    print(ans)


if __name__ == '__main__':
    main()