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;
const auto start_time = std::chrono::high_resolution_clock::now();
void timecalculater(){
    #ifndef ONLINE_JUDGE
    auto end_time = std::chrono::high_resolution_clock::now();
    std::chrono::duration<double> diff = end_time-start_time;
    cerr<<"Time Taken : "<<diff.count()<<"\n";
    #endif
}
void __print(int x) {cout << x;}
void __print(long x) {cout << x;}
void __print(long long x) {cout << x;}
void __print(unsigned x) {cout << x;}
void __print(unsigned long x) {cout << x;}
void __print(unsigned long long x) {cout << x;}
void __print(float x) {cout << x;}
void __print(double x) {cout << x;}
void __print(long double x) {cout << x;}
void __print(char x) {cout << '\'' << x << '\'';}
void __print(const char *x) {cout << '\"' << x << '\"';}
void __print(const string &x) {cout << '\"' << x << '\"';}
void __print(bool x) {cout << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cout << '{'; __print(x.first); cout << ','; __print(x.second); cout << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cout << '{'; for (auto &i: x) cout << (f++ ? "," : ""), __print(i); cout << "}";}
void _print() {cout << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cout << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define dbg(x...) cout << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define repA(i, a, n) for(int  i = a; i <= (n); ++i)
#define all(x) x.begin(), x.end()
#define sz(x) (int) (x).size()
#define fill(a,v) memset(a, v, sizeof(a))
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<long long int> vll;
typedef vector<pii> vii;
ll p=1e9+7;
const int N=2e5+5;
int main() {
   
    cin.sync_with_stdio(0); cin.tie(0);
    cin.exceptions(cin.failbit);
    int T=1;
    // cin>>T;
    while(T--){
        ll ans=0;
        ll n;cin>>n;
        vector<ll> g[n];
        map<pll,ll> ma;
        rep(i,n-1){
            ll u,v;cin>>u>>v;v--;u--;
            if(u>v)swap(u,v);
            g[u].pb(v);
            g[v].pb(u);
            ma[{u,v}]=i;
        }
        ll  m;cin>>m;
        vector<ll> res[m];
        rep(i,m){
            ll u,v;cin>>u>>v;u--;v--;
            queue<ll> q;
            ll par[n]={0};
            par[u]=-1;
            q.push(u);
            while(!q.empty()){
                ll r=q.front();
                q.pop();
                for(auto j: g[r]){
                    if(j!=par[r]){
                        par[j]=r;
                        q.push(j);
                    }
                }
            }
            while(par[v]!=-1){
                ll p=par[v];
                ll ind=ma[{min(p,v),max(p,v)}];
                res[i].pb(ind);
                v=p;
            }
        }
        ll temp=0;ans=((1<<(n-1)));
        for(ll i=1;i<(1<<m);i++){
            ll x=__builtin_popcountll(i);
            ll val=0;
            set<ll> s;
            rep(j,m){
                if(i&(1<<j)){
                    rep(k,sz(res[j]))s.insert(res[j][k]);
                }
            }
            val=1<<((n-1-sz(s)));
            if(x%2==0){
                temp-=val;
            }
            else{
                temp+=val;
            }
            // dbg(val);
        }
        cout<<ans-temp;
    }   
    timecalculater();
    return 0;   
}