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 resolve():
    def dfs(u, v, parent):
        if u == v:
            return True
        for (to, idx) in G[u]:
            if to == parent:
                continue
            if dfs(to, v, u):
                path.append(idx)
                return True
        return False

    N = int(input())
    G = [[] for _ in range(N)]
    for i in range(N-1):
        a, b = map(lambda x:int(x)-1, input().split())
        G[a].append((b, i))
        G[b].append((a, i))

    path = []

    M = int(input())
    cons = [0]*M
    # 制約ごとに、白くなる辺を入れておく
    for i in range(M):
        u, v = map(lambda x: int(x) - 1, input().split())
        path = []
        dfs(u, v, -1)
        for idx in path:
            cons[i] |= 1<<idx

    # 包除原理
    ans = 0
    for bit in range(1<<M):
        eset = 0
        for i in range(M):
            if bit>>i & 1:
                eset |= cons[i]
        white = bin(eset).count("1") # 白くなるべき辺の数
        num = 1<<(N-1-white)
        if bin(bit).count("1") % 2 == 0:
            ans += num
        else:
            ans -= num
    print(ans)

if __name__ == "__main__":
    resolve()