Problem: 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.

Examples

Input

3
1 2
2 3
1
1 3


Output

3


Input

2
1 2
1
1 2


Output

1


Input

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5


Output

9


Input

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


Output

62

Solution (in CPP):

Solution:
#include <functional>
#include <iostream>
#include <vector>

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

  long long ans = 0;
  for (int i = 0; i != 1 << m; i += 1) {
    auto res = 1LL << z[(1 << m) - 1 ^ i];
    for (int j = 0; j != m; j += 1)
      if (i >> j & 1)
        res = -res;
    ans += res;
  }
  std::cout << ans << std::endl;
}