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;
int main(){
  long N;cin>>N;
  vector<vector<pair<long,long>>>A(N);
  for(long i=0;i<N-1;i++){
    long a,b;cin>>a>>b;
    A[a-1].push_back(pair<long,long>(b-1,i));
    A[b-1].push_back(pair<long,long>(a-1,i));
  }
  long M;cin>>M;
  vector<long>U(M),V(M);
  for(long i=0;i<M;i++)cin>>U[i]>>V[i];
  
  
  vector<bitset<50>>B(M);
  for(long i=0;i<M;i++){
    queue<long>q;q.push(U[i]-1);
    vector<bitset<50>>P(N);
    while(q.size()){
      long f=q.front();q.pop();
      for(auto n:A[f])if(P[n.first].none()&&n.first!=U[i]-1){
        P[n.first]=P[f];P[n.first].set(n.second);
        q.push(n.first);
      }
    }
    B[i]=P[V[i]-1];
  }
  long ans=0;
  for(long i=0;i<(1<<M);i++){
    bitset<50>b(i),c;
    for(long j=0;j<M;j++)if(b.test(j))c|=B[j];
    ans+=(1L-((long)b.count()%2L)*2L)*(1L<<(N-1L-c.count()));  
    assert(1L<<(N-1L-c.count())>0);
  }
  cout<<ans;
  assert(ans>=0);
}