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())
E=[tuple(sorted(map(int,input().split()))) for i in range(N-1)]

DICT={x:ind for ind,x in enumerate(E)}

EDGE=[[] for i in range(N+1)]
for x,y in E:
    EDGE[x].append(y)
    EDGE[y].append(x)


M=int(input())
Q=[tuple(map(int,input().split())) for i in range(M)]

def dfs(x,y):
    USE=[-1]*(N+1)
    USE[x]=0

    NOW=[x]

    while NOW:
        n=NOW.pop()

        for to in EDGE[n]:
            if USE[to]==-1:
                USE[to]=n
                NOW.append(to)

    ANS=[y]
    while y!=x:
        y=USE[y]
        ANS.append(y)

    RET=[]

    for i in range(1,len(ANS)):
        RET.append(DICT[tuple(sorted([ANS[i-1],ANS[i]]))])

    return set(RET)

Q2=[dfs(x,y) for x,y in Q]

DP=[0]*(1<<M)
DP[0]=1

for i in range(N-1):
    B=0
    for k in range(M):
        if i in Q2[k]:
            B+=1<<k
        
    for j in range((1<<M)-1,-1,-1):
        DP[j|B]+=DP[j]

print(DP[-1])
        
    
