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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int N = 51;
vector<pair<int, int>> adj[N];

vector<int> path;
bool dfs(int u, int p, int t) {
  if (u == t) return true;
  for (auto& e: adj[u]) {
    int v = e.first;
    int i = e.second;
    if (v != p) {
      if (dfs(v, u, t)) {
        path.push_back(i);
        return true;
      }
    }
  }
  return false;
}

int main() {
  int n; cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v; cin >> u >> v;
    adj[u].emplace_back(v, i);
    adj[v].emplace_back(u, i);
  }

  int m; cin >> m;
  vector<vector<int>> paths;
  for (int i = 0; i < m; i++) {
    int u, v; cin >> u >> v;
    path.clear();
    bool res = dfs(u, u, v);
    assert(res);
    paths.push_back(path);
  }

  ll ans = 0;
  for (int mask = 0; mask < (1 << m); mask++) {
    int sign = 1;
    vector<bool> seen(n - 1, false);
    for (int i = 0; i < m; i++) {
      if (mask & (1 << i)) {
        sign *= -1;
        for (int j: paths[i]) seen[j] = true;
      }
    }
    int num = 0;
    for (bool s: seen) if (!s) num++;
    ans += sign * (1ll << num);
  }
  cout << ans << endl;

  return 0;
}
