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 PyPy3 (7.3.0) (judge verdict: Accepted)

import sys
sys.setrecursionlimit(10**7)
def main(n,ki,uv):
  # n: 頂点数
  # ki: 木
  # Euler Tour の構築
  S=[] # Euler Tour
  F=[0]*n # F[v]:vにはじめて訪れるステップ
  depth=[0]*n # 0を根としたときの深さ
  def dfs(v,pare,d):
    F[v]=len(S)
    depth[v]=d
    S.append(v)
    for i,w in ki[v]:
      if w==pare:continue
      dfs(w,v,d+1)
      S.append(v)
  dfs(0,-1,0)
  # Sをセグメント木に乗せる
  # u,vのLCAを求める:S[F[u]:F[v]+1]のなかでdepthが最小の頂点を探せば良い
  # F[u]:uに初めてたどるつくステップ
  # S[F[u]:F[v]+1]:はじめてuにたどり着いてつぎにvにたどるつくまでに訪れる頂点
  # 存在しない範囲は深さが他よりも大きくなるようにする
  INF = (n, None)
  # LCAを計算するクエリの前計算
  M = 2*n
  M0 = 2**(M-1).bit_length() # M以上で最小の2のべき乗
  data = [INF]*(2*M0)
  for i, v in enumerate(S):
    data[M0-1+i] = (depth[v], i)
  for i in range(M0-2, -1, -1):
    data[i] = min(data[2*i+1], data[2*i+2])
  # LCAの計算 (generatorで最小値を求める)
  def _query(a, b):
    yield INF
    a += M0; b += M0
    while a < b:
      if b & 1:
        b -= 1
        yield data[b-1]
      if a & 1:
        yield data[a-1]
        a += 1
      a >>= 1; b >>= 1
  # LCAの計算 (外から呼び出す関数)
  def query(u, v):
    fu = F[u]; fv = F[v]
    if fu > fv:
      fu, fv = fv, fu
    return S[min(_query(fu, fv+1))[1]]
  
  todo=[[0,[]]]
  path=[[]]*n
  mi=[1]*n
  while todo:
    v,ary=todo.pop()
    path[v]=ary
    mi[v]=0
    for i,nv in ki[v]:
      if mi[nv]==1:
        todo.append([nv,ary+[i]])
        mi[nv]=1

  for i in range(n):
    path[i]=set(path[i])
  memo={}
  ans=0
  for i in range(2**m):
    chk=set(())
    for j in range(m):
      if (i>>j)&1:
        chk.add(j)
    must_w=set(())
    for c in chk:
      if c in memo:
        tmp=memo[c]
      else:
        u,v=uv[c]
        dd=query(u,v)
        tmp=path[u]|path[v]
        tmp-=path[dd]
        memo[c]=tmp
      must_w|=tmp
    t=n-1-len(must_w)
    if len(chk)%2==1:
      ans-=pow(2,t)
    else:
      ans+=pow(2,t)
  print(ans)
if __name__=='__main__':
  n=int(input())
  ki=[[] for _ in range(n)]
  for i in range(n-1):
    u,v=map(int,input().split())
    u,v=u-1,v-1
    ki[u].append([i,v])
    ki[v].append([i,u])
  m=int(input())
  uv=[]
  for _ in range(m):
    u,v=map(int,input().split())
    u,v=u-1,v-1
    uv.append([u,v])
  main(n,ki,uv)