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: Accepted)

#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define each(x, y) for (auto &x : y)
#define all(a) (a).begin(), (a).end()
#define bit(b) (1ll << (b))

using namespace std;

using i32 = int;
using i64 = long long;
using u64 = unsigned long long;
using f80 = long double;
using vi32 = vector<i32>;
using vi64 = vector<i64>;
using vu64 = vector<u64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << '\n'; exit(0); }
inline void no() { cout << "No" << '\n'; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { return a / gcd(a, b) * b; }
inline u64 xorshift() { static u64 x = 88172645463325252ull; x = x ^ (x << 7); return x = x ^ (x >> 9); }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> inline void amax(T &x, T y) { if (x < y) x = y; }
template <typename T> inline void amin(T &x, T y) { if (x > y) x = y; }
template <typename T> inline T exp(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> inline T bis(T ok, T ng, function<bool(T)> f, T eps = 1) { while (abs(ok - ng) > eps) { T mi = (ok + ng) / 2; (f(mi) ? ok : ng) = mi; } return ok; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { each(x, v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
template <typename T, typename S> istream& operator>>(istream &is, pair<T, S> &p) { is >> p.first >> p.second; return is; }
template <typename T, typename S> ostream& operator<<(ostream &os, pair<T, S> &p) { os << p.first << ' ' << p.second; return os; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }

void solve() {
  int n;
  cin >> n;
  vector<vector<pair<int, int>>> edges(n);
  rep(i, n - 1) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    edges[a].emplace_back(b, i);
    edges[b].emplace_back(a, i);
  }
  int m;
  cin >> m;
  vi32 es;
  function<bool(int, int, int)> dfs = [&](int u, int p, int end) {
    if (u == end) {
      es = vi32();
      return true;
    }
    each(pa, edges[u]) {
      int v, e;
      tie(v, e) = pa;
      if (v == p) continue;
      if (dfs(v, u, end)) {
        es.emplace_back(e);
        return true;
      }
    }
    return false;
  };
  vi64 shu(m);
  rep(i, m) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    dfs(u, -1, v);
    each(e, es) shu[i] |= 1ll << e;
  }
  i64 ans = exp(2ll, n - 1);
  reps(i, (1 << m) - 1) {
    i64 se = 0;
    rep(j, m) {
      if (i & (1 << j)) se |= shu[j];
    }
    int rest = n - 1 - __builtin_popcountll(se);
    ans += exp(2ll, rest) * (__builtin_parityll(i) ? -1 : 1);
  }
  cout << ans << endl;
}