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:
'''
包除原理
すべての条件を満たしている
2^{N-1}から「どれかの条件を満たしていない」を引けば良い
i番目の条件を満たしていない確率はかんたんに計算できる
Sに属していいないやつが全て条件を満たしていない確率については
Sに対応する辺が全て白となる確率を考えれば良い
'''
from collections import deque
N=int(input())
G=[[] for i in range(N)]
a=[0 for i in range(N-1)]
b=[0 for i in range(N-1)]
edges=dict()
for i in range(N-1):
    a[i],b[i]=map(int,input().split())
    a[i]-=1;b[i]-=1
    G[a[i]].append(b[i])
    G[b[i]].append(a[i])
    edges[(min(a[i],b[i]),max(a[i],b[i]))]=i
M=int(input())
u=[0 for i in range(M)]
v=[0 for i in range(M)]
for i in range(M):
    u[i],v[i]=map(int,input().split())
    u[i]-=1;v[i]-=1
root=[0 for i in range(M)]
for i in range(M):
    q=deque([u[i]])
    parents=[-1 for i in range(N)]
    reached=[0 for i in range(N)]
    while(len(q)>0):
        r=q.popleft()
        reached[r]=1
        for p in G[r]:
            if reached[p]==1:
                continue
            parents[p]=r
            q.append(p)
    x=v[i]
    while(1):
        if parents[x]==-1:
            break
        s=min(x,parents[x])
        t=max(x,parents[x])
        root[i]+=2**edges[(s,t)]
        x=parents[x]
ans=0
def cnt(x):
    ans=0
    for i in range(M):
        if x|(2**i)==x:
            ans+=1
    return ans
q=deque([])
q.append(0)
S=[0 for i in range(2**M)]
reached=[0 for i in range(2**M)]
while(len(q)>0):
    i=q.popleft()
    for j in range(M):
        if i&(2**j)==0:
            S[i|(2**j)]=S[i]|root[j]
            if reached[i|(2**j)]==0:
                q.append(i|(2**j))
                reached[i|(2**j)]=1
    ans+=(-1)**(cnt(i))*2**(N-bin(S[i]).count("1")-1)
print(ans)