Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

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

Sample Output 4

62

The tree in this input is shown below:

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

def get_paths(start, goal, g, pre)
  return false if @reached
  #ゴールなら終了
  if start == goal
    @reached = true
    return true
  end
  #startからつながる辺を検索する
  g[start].each do |p|
    #元の頂点に戻るパスならスキップ
    next if p.first == pre
    #つながる頂点を引数にして再帰
    if get_paths(p.first, goal, g, start)
      #ゴールしている経路なら辺を記録する
      bit = 1 << p[1] #辺をbitフラグで管理( 10 は1番目の辺を使っている
      @path |= bit
      return true
    end
  end

  #ゴールできなかったら終了
  return false
end

def main
  n = gets.chomp.to_i
  graph = []

  (n-1).times do |i|
    a, b = gets.chomp.split.map(&:to_i)
    
    graph[a] = [] if !graph[a]
    graph[b] = [] if !graph[b]
    graph[a].push([b, i])
    graph[b].push([a, i])
  end

  m = gets.chomp.to_i
  paths = []
  m.times do |i|
    @reached = false
    @path = 0
    x, y = gets.chomp.split.map(&:to_i)
    get_paths(x, y, graph, -1)
    paths.push(@path)
  end

  ans = 0
  #回答の導出
  (1 << m).times do |i|
    side_cnt = 0 #制約される辺を表す( < n - 1)
    bit = i.to_s(2).chars.reverse #使用する制約を表す(eg. 101)
    bit.length.times do |i|
      if bit[i] == '1'
        side_cnt |= paths[i]
      end
    end

    k = side_cnt.to_s(2).count('1')
    j = i.to_s(2).count('1')
    if j % 2 == 0
      ans += 2 ** (n - 1 - k)
    else
      ans -= 2 ** (n - 1 - k)
    end
  end
  puts ans
end

main