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 (Clang 3.8.0) (judge verdict: Accepted)

#include <cstdio>
#include <vector>

#include <cstdint>

constexpr bool parity32(std::uint_fast32_t c) noexcept {
  c ^= c >> 1;
  c ^= c >> 2;
  return ((c & 0x11111111) * 0x11111111 >> 28 & 0x1) != 0;
}

int scan() {
  int r = 0;
  int c;
  while ((c = getchar_unlocked() - '0') >= 0)
    r = r * 10 + c;
  return r;
}

int main() {
  const int n = scan();
  std::vector<std::vector<int>> t(n);
  for (int i = 0; i != n - 1; i += 1) {
    const int a = scan() - 1;
    const int b = scan() - 1;
    t[a].push_back(b);
    t[b].push_back(a);
  }
  const int m = scan();
  const int s = 1 << m;
  std::vector<int> z(s, 0);
  {
    std::vector<int> x(n, 0);
    for (int i = 0; i != m; i += 1) {
      const int u = scan() - 1;
      const int v = scan() - 1;
      x[u] ^= 1 << i;
      x[v] ^= 1 << i;
    }
    const auto dfs = [&](const auto &dfs, int v, int p) -> void {
      for (int e : t[v]) {
        if (e != p) {
          dfs(dfs, e, v);
          z[x[e]] += 1;
          x[v] ^= x[e];
        }
      }
    };
    dfs(dfs, 0, n);
  }
  for (int w = 1; w != s; w <<= 1) {
    for (int i = 0; i != s; i += w << 1) {
      const int end = i + w;
      for (int j = i; j != end; j += 1) {
        z[j + w] += z[j];
      }
    }
  }

  long long ans = 0;
  for (int i = 0; i != 1 << m; i += 1) {
    auto res = 1LL << z[s - 1 ^ i];
    if (parity32(i))
      res = -res;
    ans += res;
  }
  printf("%lld", ans);
}