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 os
import sys
from functools import lru_cache

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353

N = int(sys.stdin.buffer.readline())
AB = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N - 1)]
M = int(sys.stdin.buffer.readline())
VU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(M)]

# N <= 50
# M <= 20


graph = [[] for _ in range(N)]
edges = {}
for i, (a, b) in enumerate(AB):
    a -= 1
    b -= 1
    graph[a].append(b)
    graph[b].append(a)
    edges[a, b] = i
    edges[b, a] = i


@lru_cache(maxsize=None)
def path(v, u, p):
    if v == u:
        return [u]
    for nv in graph[u]:
        if nv == p:
            continue
        ret = path(v, nv, u)
        if ret is not None:
            return ret + [u]
    return None


# 辺ごとに、黒くしたときにどの制約を満たすことができるかを持つ
constraints = [0 for _ in range(N - 1)]
for i, (v, u) in enumerate(VU):
    v -= 1
    u -= 1
    pp = path(v, u, None)
    for pv, pu in zip(pp, pp[1:]):
        constraints[edges[pv, pu]] |= 1 << i

# for c in constraints:
#     import numpy as np
#
#     print(np.binary_repr(c, M))

# or したときに全部 True になる組み合わせの数
dp = [0 for _ in range(1 << M)]
dp[0] = 1
for c in constraints:
    for i in reversed(range(1 << M)):
        dp[i | c] += dp[i]
print(dp[-1])
