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;

vector<pair<int, int>> tree[50];
vector<unsigned int> paths(49);

bool dfs_mem(long c, long v, long p, long r) {
  if (c == v)
    return true;
  for (auto t : tree[c]) {
    if (t.second != p and dfs_mem(t.second, v, c, r)) {
      paths[t.first] ^= 1ull << r;
      return true;
    }
  }
  return false;
}

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int N; cin>>N;
  for (int i = 0; i < N-1; i++) {
    int a,b; cin>>a>>b;
    a--; b--;
    tree[a].emplace_back(i, b);
    tree[b].emplace_back(i, a);
  }
  int M; cin>>M;
  for (int i = 0; i < M; i++) {
    int u,v; cin>>u>>v;
    u--; v--;
    dfs_mem(u, v, -1, i);
  }

  vector<long> dp(1ull << M);
  dp[0] = 1;
  for (int i = 0; i < N-1; i++) {
    for (int j = (1ull<<M)-1; j >= 0; j--) {
      dp[j | paths[i]] += dp[j];
    }
  }
  cout << dp[(1ull<<M)-1] << endl;

  return 0;
}
