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
N=int(input())
edge={i:[] for i in range(1,N+1)}
for i in range(0,N-1):
    a,b=map(int,input().split())
    edge[a].append((b,i))
    edge[b].append((a,i))

parent={i:-1 for i in range(1,N+1)}
depth={i:-1 for i in range(1,N+1)}
parent[1]=(0,-1)
depth[1]=0
q=set([1])
sub=set([])
while q:
    while q:
        x=q.pop()
        for b,i in edge[x]:
            if parent[b]==-1:
                parent[b]=(x,i)
                depth[b]=depth[x]+1
                sub.add(b)

        if not q:
            q=sub
            sub=set([])
            break

M=int(input())
condition=[[] for i in range(0,M)]
for i in range(0,M):
    u,v=map(int,input().split())
    ans=set([])
    while depth[u]!=depth[v]:
        if depth[u]>depth[v]:
            U,j=parent[u]
            ans.add(j)
            u=U
        else:
            V,j=parent[v]
            ans.add(j)
            v=V

    while u!=v:
        U,j=parent[u]
        ans.add(j)
        u=U
        V,j=parent[v]
        ans.add(j)
        v=V
    con=0
    for j in ans:
        con+=2**j
    condition[i]=con

ans=2**(N-1)
ban=0
for i in range(1,2**M):
    num=bin(i).count('1')
    s=0
    for j in range(0,M):
        if i>>j &1==1:
            s=s|condition[j]


    n=bin(s).count('1')
    if num%2==1:
        ban+=2**(N-1-n)
    else:
        ban-=2**(N-1-n)

print(ans-ban)