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())
AB = [tuple(map(int,input().split())) for i in range(N-1)]
M = int(input())
UV = [tuple(map(int,input().split())) for i in range(M)]

es = [[] for i in range(N)]
for i,(a,b) in enumerate(AB):
    a,b = a-1,b-1
    es[a].append((b,i))
    es[b].append((a,i))

paths = [0] * M
for i,(u,v) in enumerate(UV):
    u,v = u-1,v-1
    stack = [u]
    visited = [0] * N
    visited[u] = 1
    edges = [-1] * N
    while stack:
        s = stack.pop()
        for to,j in es[s]:
            if visited[to]: continue
            visited[to] = 1
            edges[to] = j
            stack.append(to)
            if to == v:
                break
        else:
            continue
        while edges[v] >= 0:
            j = edges[v]
            paths[i] |= (1<<j)
            a,b = AB[j]
            a,b = a-1,b-1
            v = a if a!=v else b
        break

dp = [0] * (1<<M)
for b in range(1,1<<M):
    k = len(bin(b))-3
    dp[b] = dp[b-2**k] | paths[k]

ans = 0
for i,d in enumerate(dp):
    b = N-1 - bin(d).count('1')
    ans += 2**b if bin(i).count('1')%2==0 else -2**b
print(ans)