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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
int r[200001]={0},d[200001]={0},e=0,f=0;
vector<int> G[200001];
map<P,int> A;
void dfs(int x,int y,int z=-1){
    if(x==y){
        f=1;
        return;
    }
    for(auto i:G[x]){
        if(i==z) continue;
        if(f==1) continue;
        dfs(i,y,x);
        if(f==1) A[P(min(i,x),max(i,x))]+=1<<e;
    }
}
int main(){
    ll n,m,a[50],b[50],u,v,B[50],ans=0;
    cin>>n;
    rep(i,n-1){
        cin>>a[i]>>b[i];
        G[a[i]].push_back(b[i]);
        G[b[i]].push_back(a[i]);
    }
    cin>>m;
    rep(i,m) cin>>u>>v,dfs(u,v),e++,f=0;
    rep(i,n-1) B[i]=A[P(min(a[i],b[i]),max(a[i],b[i]))];
    for(int i=1;i<(1<<m);i++){
        ll num=0,x=0;
        for(int j=i;j!=0;j>>=1) num+=j&1;
        rep(k,n-1){
            int z=1;
            rep(j,m){
                if(i>>j&1){
                    if(B[k]>>j&1) z=0;
                }
            }
            if(z) x++;
        }
        if(num%2) ans+=pow(2,x);
        else ans-=pow(2,x);
    }
    cout<<pow(2,n-1)-ans<<"\n";
}