Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

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

Sample Output 4

62

The tree in this input is shown below:

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

template<typename F>
struct FixPoint : F{
  FixPoint(F&& f):F(forward<F>(f)){}
  template<typename... Args>
  decltype(auto) operator()(Args&&... args) const{
    return F::operator()(*this,forward<Args>(args)...);
  }
};
template<typename F>
inline decltype(auto) MFP(F&& f){
  return FixPoint<F>{forward<F>(f)};
}

//INSERT ABOVE HERE
signed main(){
  int n;
  cin>>n;

  vector< vector<int> > G(n);
  using P = pair<int, int>;
  map<P, int> idx;
  for(int i=1;i<n;i++){
    int a,b;
    cin>>a>>b;
    a--;b--;
    G[a].emplace_back(b);
    G[b].emplace_back(a);
    idx[minmax({a,b})]=i;
  }

  int m;
  cin>>m;
  vector<int> us(m),vs(m);
  for(int i=0;i<m;i++) cin>>us[i]>>vs[i],us[i]--,vs[i]--;

  vector< vector<int> > par(n,vector<int>(n,-1));
  for(int i=0;i<n;i++){
    MFP([&](auto dfs,int v,int p)->void{
          par[i][v]=p;
          for(int u:G[v]){
            if(u==p) continue;
            dfs(u,v);
          }
        })(i,-1);
  }

  using ll = long long;
  vector<ll> po(n,1);
  for(int i=0;i+1<n;i++) po[i+1]=po[i]*2;

  vector<ll> es(m,0);
  for(int i=0;i<m;i++){
    int x=us[i],y=vs[i];
    while(x!=y){
      es[i]|=1LL<<idx[minmax({x,par[y][x]})];
      x=par[y][x];
    }
  }

  int sz=1<<m;
  ll ans=0;
  for(int bit=0;bit<sz;bit++){
    ll res=0;
    for(int i=0;i<m;i++)
      if((bit>>i)&1) res|=es[i];

    if(__builtin_parity(bit))
      ans-=po[n-1-__builtin_popcountll(res)];
    else
      ans+=po[n-1-__builtin_popcountll(res)];
  }

  cout<<ans<<endl;
  return 0;
}