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
def main():
    import sys
    def I():
        return int(sys.stdin.readline().rstrip())

    def MI():
        return map(int, sys.stdin.readline().rstrip().split())

    N = I()
    dist = [[100] * (N + 1) for _ in range(N + 1)]  # 2点間の最短距離
    edges = {}

    for i in range(N - 1):
        a, b = MI()
        dist[a][b] = 1
        dist[b][a] = 1
        edges[(a, b)] = i
        edges[(b, a)] = i

    for i in range(1, N + 1):
        dist[i][i] = 0

    for k in range(1, N + 1):
        for i in range(1, N + 1):
            for j in range(1, N + 1):
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])

    M = I()
    A = []  # uとvを結ぶ経路上にある辺の集合
    for _ in range(M):
        u, v = MI()
        B = set()
        while True:
            if dist[u][v] == 1:
                B.add(edges[(u, v)])
                break
            else:
                for i in range(1, N + 1):
                    if dist[u][i] == dist[u][v] - 1 and dist[i][v] == 1:
                        B.add(edges[(i, v)])
                        v = i
                        break
        A.append(B)

    # 包除原理

    ans = 0
    for i in range(2 ** M):
        C = set()
        a = 0
        for j in range(M):
            if (i >> j) & 1:
                C |= A[j]
                a += 1
        l = len(C)
        a %= 2
        ans += (-1) ** a * 2 ** (N - 1 - l)

    print(ans)

if __name__ == '__main__':
    main()
