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 copy import copy

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

edge_set = [set() for i in range(n)]
edge_id = 0

def dfs(p,v,se):
    global edge_id
    edge_set[v] = se
    for to in G[v]:
        if to != p:
            nset = copy(se)
            nset.add(edge_id)
            edge_id += 1
            dfs(v,to,nset)

dfs(-1,0,set())

m = int(input())
edge_set2 = []
for i in range(m):
    a,b = map(int, input().split())
    a,b = a-1,b-1
    edge_set2.append(edge_set[a]^edge_set[b])

from itertools import combinations
ans = 0
for i in range(m):
    op = -1 if i%2 else 1
    for c in combinations(range(m), i+1):
        s = set()
        for j in c:
            s |= edge_set2[j]
        rem = n-1 - len(s)
        ans += op*(1<<rem)
ans = 2**(n-1) - ans
print(ans)

