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 <iostream>
#include <vector>
using namespace std;

int main(){
  int N;
  const int INF = 1e9;
  cin >> N;
  vector<vector<int>> D(N,vector<int>(N,INF));
  vector<pair<int,int>> E;
  for(int i = 1; i < N; ++i){
    int a, b;
    cin >> a >> b;
    --a,--b;
    D[a][b] = 1;
    D[b][a] = 1;
    E.emplace_back(a,b);
  }
  for(int i = 0; i < N; ++i) D[i][i] = 0;
  for(int k = 0; k < N; ++k){
    for(int i = 0; i < N; ++i){
      for(int j = 0; j < N; ++j){
        D[i][j] = min(D[i][j],D[i][k]+D[k][j]);
      }
    }
  }
  int M;
  cin >> M;
  vector<pair<int,int>> uv;
  for(int i = 0; i < M; ++i){
    int u, v;
    cin >> u >> v;
    --u,--v;
    uv.emplace_back(u,v);
  }
  long long ans = 0;
  for(int i = 0; i < (1<<M); ++i){
    int n = N-1;
    for(auto e : E){
      bool f = false;
      for(int j = 0; j < M; ++j){
        if(not (i>>j&1)) continue;
        auto x = uv[j];
        if((D[x.first][x.second] == D[x.first][e.first]+1+D[e.second][x.second]) or
           (D[x.first][x.second] == D[x.first][e.second]+1+D[e.first][x.second])){
          f = true;
          break;
        }
      }
      if(f) --n;
    }
    if(__builtin_popcount(i)%2) ans -= 1LL<<n;
    else ans += 1LL<<n;
  }
  cout << ans << endl;
}
