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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define fi first
#define se second
using namespace std;
using P = pair<long long, long long>;

struct data {
  int to, n;
};

long long n, m;
vector<long long> dp;
vector<vector<data>> edge;
vector<P> v;
vector<vector<vector<long long>>> lst;

long long twopow(long long x) {
  long long res = 1, now = 2;
  while(x) {
    if(x & 1) res *= now;
    now *= now;
    x >>= 1;
  }
  return res;
}

unsigned long long popcount(unsigned long long x) {
  x = ((x & 0xaaaaaaaaaaaaaaaaUL) >> 1) +
      (x & 0x5555555555555555UL);
  x = ((x & 0xccccccccccccccccUL) >> 2) +
      (x & 0x3333333333333333UL);
  x = ((x & 0xf0f0f0f0f0f0f0f0UL) >> 4) +
      (x & 0x0f0f0f0f0f0f0f0fUL);
  x = ((x & 0xff00ff00ff00ff00UL) >> 8) +
      (x & 0x00ff00ff00ff00ffUL);
  x = ((x & 0xffff0000ffff0000UL) >> 16) +
      (x & 0x0000ffff0000ffffUL);
  x = ((x & 0xffffffff00000000UL) >> 32) +
      (x & 0x00000000ffffffffUL);
  return x;
}

long long solve();

int main() {
  cin >> n;
  edge.resize(n);
  for(int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    edge[a].push_back({b, i});
    edge[b].push_back({a, i});
  }
  cin >> m;
  v.resize(m);
  for(int i = 0; i < m; ++i) {
    cin >> v[i].fi >> v[i].se;
    --v[i].fi, --v[i].se;
  }
  cout << solve() << endl;
  return 0;
}

long long solve() {
  long long res = 0;
  lst.resize(n, vector<vector<long long>>(n));
  for(int i = 0; i < n; ++i) {
    queue<int> qu;
    vector<bool> used(n, 0);
    qu.push(i);
    used[i] = 1;
    while(qu.size()) {
      int now = qu.front();
      qu.pop();
      for(auto e : edge[now]) {
        int nextp = e.to;
        if(!used[nextp]) {
          lst[i][nextp] = lst[i][now];
          lst[i][nextp].push_back(e.n);
          used[nextp] = 1;
          qu.push(nextp);
        }
      }
    }
  }

  for(long long i = 0; i < (1LL << m); ++i) {
    long long now = 0;
    set<int> st;
    for(int j = 0; j < m; ++j)
      if(i >> j & 1) {
        for(auto id : lst[v[j].fi][v[j].se]) st.insert(id);
      }
    now = twopow(n - 1 - st.size());
    if(popcount(i) % 2)
      res -= now;
    else
      res += now;
  }
  return res;
}