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
from collections import deque


def dfs(start, goal, tree):
    """startからgoalまでの木上のpathを出力する"""
    q = deque([start])
    visited = {}
    visited[start] = -1
    while True:
        pos = q.pop()
        if pos == goal:
            break
        for child in tree[pos]:
            if child in visited:
                continue
            visited[child] = pos
            q.append(child)
    res = []
    pos = goal
    while True:
        par = visited[pos]
        if par == -1:
            break
        res.append((min(par, pos), max(par, pos)))
        pos = par
    return res


n = int(input())
info = [list(map(int, input().split())) for i in range(n - 1)]
m = int(input())
path = [list(map(int, input().split())) for i in range(m)]

tree = [[] for i in range(n)]
memo = {}
for i in range(n - 1):
    a, b = info[i]
    a -= 1
    b -= 1
    tree[a].append(b)
    tree[b].append(a)
    memo[(min(a, b), max(a, b))] = i
  

li_path = [0] * m
for i in range(m):
    s, g = path[i]
    s -= 1
    g -= 1
    res = dfs(s, g, tree)
    tmp = 0
    for j in res:
        tmp += (1 << memo[j])
    li_path[i] = tmp
  
ans = 0
for bit_state in range(1 << m):
    mask = 0
    cnt = 0
    for i in range(m):
        if bit_state & (1 << i):
            cnt += 1
            mask |= li_path[i]
    num_n = bin(mask).count("1")
    if cnt % 2 == 0:
        ans += 2 ** (n - 1 - num_n)
    else:
        ans -= 2 ** (n - 1 - num_n)
print(ans)
