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

readline = sys.stdin.readline
MOD = 10 ** 9 + 7
INF = float('INF')
sys.setrecursionlimit(10 ** 5)


def main():
    from collections import defaultdict, deque
    from itertools import combinations

    N = int(readline())

    edge = defaultdict(list)

    for i in range(N - 1):
        a, b = map(int, readline().split())
        edge[a - 1].append((b - 1, i))
        edge[b - 1].append((a - 1, i))

    def bfs(s, g):
        que = deque()
        que.append((s, []))
        visited = [False] * N
        visited[s] = True

        while que:
            cur, path = que.popleft()
            if cur == g:
                return set(path)
            for nx, i in edge[cur]:
                if not visited[nx]:
                    new_path = path + [i]
                    que.append((nx, new_path))
                    visited[nx] = True

    M = int(readline())
    constraints = []
    for _ in range(M):
        u, v = map(int, readline().split())
        u -= 1
        v -= 1
        constraints.append(bfs(u, v))

    cnt = 0
    for i in range(1, M + 1):
        sign = 1 if i % 2 == 1 else -1
        for per in combinations(range(M), r=i):
            incl = set()
            for x in per:
                incl = incl.union(constraints[x])
            cnt += sign * 2 ** ((N - 1) - len(incl))

    print(2 ** (N - 1) - cnt)


if __name__ == '__main__':
    main()
