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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; ++i)
#define all(a) (a).begin(), (a).end()

int N;

int main() {
  cin >> N;
  vector<vector<int>> G(N);
  vector<bitset<20>> C(N - 1, 0);
  unordered_map<int, unordered_map<int, int>> edge;
  rep(i, N - 1) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    G[a].push_back(b);
    G[b].push_back(a);
    edge[a][b] = i;
    edge[b][a] = i;
  }

  int M;
  cin >> M;

  function<bool(int, int, int, int)> append = [&](int u, int v, int p, int c) {
    if (u == v) return true;

    bool found = false;
    for (auto w : G[u]) {
      if (w == p) continue;
      bool goodPath = append(w, v, u, c);
      if (goodPath) C[edge[u][w]].set(c);
      found |= goodPath;
    }
    return found;
  };

  rep(i, M) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    append(u, v, -1, i);
  }

  vector<vector<ll>> dp((1 << M), vector<ll>(N, 0));
  dp[0][0] = 1;
  rep(i, N - 1) rep(s, (1 << M)) {
    dp[s | int(C[i].to_ulong())][i + 1] += dp[s][i];  // B
    dp[s][i + 1] += dp[s][i];                         // W
  }

  cout << dp[(1 << M) - 1][N - 1] << endl;

  return 0;
}
