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 itertools import combinations

N = int(input())

adj = [list() for _ in range(N)]
for _ in range(N-1):
    a,b = map(int,input().split())
    adj[a-1].append(b-1)
    adj[b-1].append(a-1)

parent = [None]*N
visited = [False]*N
visited[0] = True
def rec(u):
    for v in adj[u]:
        if not visited[v]:
            parent[v] = u
            visited[v] = True
            rec(v)
rec(0)
del adj

M = int(input())


# edge_id to crucial query_id
tasks = []

for q in range(M):
    u,v = map(int,input().split())
    u,v = u-1,v-1
    u2r = []
    v2r = []
    while u is not None:
        u2r.append(u)
        u = parent[u]
    while v is not None:
        v2r.append(v)
        v = parent[v]
    
    for i,(a,b) in enumerate(zip(reversed(u2r), reversed(v2r))):
        if a != b:
            break
    else:
        i += 1
    # path = u2r[:-i] + [u2r[-i]] + list(reversed(v2r[:-i]))
    tasks.append(u2r[:-i]+v2r[:-i])

cnt = 0
sign = 1

for n in range(1,M+1):
    sub = 0
    for p in combinations(range(M),n):
        sub += 2**(N-1-len(set().union(*(tasks[i] for i in p))))
    cnt += sign*sub
    sign *= -1
print(2**(N-1)-cnt)