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
import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7 # 998244353
input=lambda:sys.stdin.readline().rstrip()
def resolve():
    n = int(input())
    E = [[] for _ in range(n)]
    edge_to_idx = {}
    for i in range(n - 1):
        u, v = map(int,input().split())
        u -= 1
        v -= 1
        E[u].append(v)
        E[v].append(u)
        edge_to_idx[u, v] = edge_to_idx[v, u] = i

    def path(v, p, target):
        # base case
        if v == target:
            return 1 << edge_to_idx[v, p]
        res = 0
        for nv in E[v]:
            if nv == p:
                continue
            res |= path(nv, v, target)
        if res > 0 and p != -1:
            res |= 1 << edge_to_idx[v, p]
        return res

    m = int(input())
    constraints = [None] * m
    for i in range(m):
        u, v = map(int,input().split())
        u -= 1
        v -= 1
        constraints[i] = path(u, -1, v)

    res = 1 << (n - 1)
    # inc - exc principle
    for U in range(1, 1 << m):
        coef = 1 if bin(U).count('1') & 1 else -1
        binary = 0
        for i in range(m):
            if (U >> i) & 1:
                binary |= constraints[i]
        cnt = (n - 1) - bin(binary).count('1')
        res -= coef * (1 << cnt)

    print(res)
resolve()