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 deque


def getPath(adj, start, goal):
    q = deque()
    q.appendleft((start, 0))
    while len(q):
        cur, p = q.pop()
        if cur == goal:
            return p
        for nxt, nhop in adj[cur]:
            if 2**nhop & p:
                continue
            npath = p | 2**nhop
            q.appendleft((nxt, npath))


def bitcount(x):
    return bin(x).count("1")


def main():
    N = int(input())
    adj = [set() for _ in range(N)]
    for i in range(N-1):
        a, b = map(int, input().split())
        adj[a-1].add((b-1, i))
        adj[b-1].add((a-1, i))
    M = int(input())
    dp = [0] * (2**M)
    for i in range(M):
        u, v = map(int, input().split())
        path = getPath(adj, u-1, v-1)
        for j in range(2**i):
            dp[j + 2**i] = dp[j] | path
    ans = 0
    for i in range(2**M):
        ans += (-1) ** bitcount(i) * 2 ** (N - bitcount(dp[i]) - 1)
    print(ans)


if __name__ == "__main__":
    main()
