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:
from collections import deque

def main():
    n = int(input())
    edge = []
    edge_id = []
    for i in range(n):
        edge.append([])
        edge_id.append(dict())
    for i in range(n-1):
        a, b = map(int, input().split())
        edge[a-1].append(b-1)
        edge[b-1].append(a-1)
        edge_id[a-1].update({b-1: i})
        edge_id[b-1].update({a-1: i})
    m = int(input())
    uv = []
    for i in range(m):
        uv.append(list(map(int, input().split())))

    edge_mask = []
    for i in range(m):
        distance_from_u = []
        visited = []
        for j in range(n):
            distance_from_u.append(0)
            visited.append(False)
        que = deque()
        que.append([uv[i][0]-1, 0])
        while len(que) > 0:
            now, dist = que.popleft()
            if visited[now]:
                continue
            visited[now] = True
            distance_from_u[now] = dist
            for x in edge[now]:
                if not visited[x]:
                    que.append([x, dist+1])
        next = uv[i][1]-1
        edge_set = set()
        while distance_from_u[next] > 0:
            for x in edge[next]:
                if distance_from_u[x] < distance_from_u[next]:
                    edge_set.add(edge_id[next][x])
                    next = x
                    break
        edge_mask.append(edge_set)

    dp_mask = []
    for i in range(n-1):
        new_mask = 0
        for j in range(m):
            if i in edge_mask[j]:
                new_mask = new_mask | (1 << j)
        dp_mask.append(new_mask)

    dp = []
    for i in range(n):
        dp.append([])
        for j in range(1 << m):
            dp[i].append(0)
    dp[0][0] = 1
    for i in range(n-1):
        for j in range(1 << m):
            dp[i+1][j] += dp[i][j]
            dp[i+1][j|dp_mask[i]] += dp[i][j]

    ans = dp[n-1][(1<<m)-1]

    print(ans)

if __name__ == '__main__':
    main()