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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<cstdint>

int const MAXN = 50;
int const MAXM = 20;
int N, M;

std::vector<int> G[MAXN];
std::pair<int, int> diff[MAXN][3];
int tmp[MAXN];
std::set<int> s;

bool dfs(int v, int p, int u) {
  bool res = (v==u);
  if(!res) for(auto c: G[v]) if(c != p) {
    res = dfs(c, v, u);
    if(res) break;
  }
  if(p >= 0 && res) s.insert(p*MAXN+v);
  return res;
}

void dfs2(int v, int p) {
  if(s.count(p*MAXN+v)+s.count(v*MAXN+p) > 0) {
    --tmp[p]; ++tmp[v];
  }
  for(auto c: G[v]) if(c != p) dfs2(c, v);
}

void calc(int v, int u, int ci) {
  s.clear();
  dfs(u, -1, v);
  std::fill_n(tmp, N, 0);
  dfs2(0, -1);
  diff[ci][2].first = MAXN;
  int cp = 0;
  for(int i = 0; i < N; ++i) if(tmp[i]!=0) {
    diff[ci][cp++] = std::make_pair(i, tmp[i]);
  }
}

std::pair<int, int> sum_cond(int v, int p) {
  std::pair<int, int> res(0, 0);
  for(auto c: G[v]) if(c != p) {
    auto pa = sum_cond(c, v);
    res.first += pa.first;
    res.second += pa.second;
  }
  res.second += tmp[v];
  if(res.second > 0) ++res.first;
  return res;
}

int main() {
  std::cin >> N;
  for(int i = 0; i < N-1; ++i) {
    int a, b; std::cin >> a >> b; --a; --b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  std::cin >> M;
  for(int i = 0; i < M; ++i) {
    int u, v; std::cin >> u >> v; --u; --v;
    calc(v, u, i);
  }
  
  int64_t ans = 0;
  for(int s = 0; s < 1<<M; ++s) {
    int64_t i = 1;
    std::fill_n(tmp, N, 0);
    for(int j = 0; j < M; ++j) if(s>>j&1) {
      for(int k = 0; k < 3; ++k) if(diff[j][k].first < N) {
        auto& pa = diff[j][k];
        tmp[pa.first] += pa.second;
      }
      i *= -1;
    }
    ans += i*(1 << N-1-sum_cond(0, -1).first);
  }
  std::cout << ans << std::endl;
  return 0;
}