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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

#define IOS ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define I insert
#define endl '\n'
#define all(x) x.begin(), x.end()

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
 
using vvll = vector< vector<ll> >;
using vll = vector<ll> ;
using vb = vector<bool> ;
using pll = pair<ll, ll> ;

const ll maxs = 1e6+5;
const ll lmaxs = 20;
ll mod = 1e9+7;
ll oo = 1e15;

ll fast(ll x,ll y){
    ll res=1;
    while(y){
        if(y&1)
            res = (res*x)%mod;
        x = (x*x)%mod;
        y >>= 1;
    }
    return res;
}

ll inv(ll n){
    return fast(n,mod-2);
}


void pre(){


}

vector<vector<int>> tr,par;
vector<int> lvl;
vector<int> prefix;

void dfs(int node,int parent){
    par[node][0] = parent;
    for(int i=1;par[node][i-1] != -1 ;i++){
        par[node][i] = par[par[node][i-1]][i-1];
    }
    for(auto child : tr[node]){
        if(child != parent){
            lvl[child] = lvl[node] + 1;
            dfs(child,node);
        }
    }
}

ll lca(ll src,ll dest){
    if(lvl[src] < lvl[dest])
        swap(src,dest);
    for(int i=19;i>=0;i--){
        if(par[src][i] != -1 and lvl[par[src][i]] >= lvl[dest]){
            src = par[src][i];
        }
    }

    if(src==dest)
        return src;
    
    for(int i=19;i>=0;i--){
        if(par[src][i] != par[dest][i]){
            src = par[src][i];
            dest = par[dest][i];
        }
    }
    return par[src][0];
}

void dfs1(int node,int parent){

    for(auto child : tr[node]){
        if(child != parent){
            dfs1(child,node);
            prefix[node] += prefix[child]; 
        }
    }

}

void solve(){

    int n;
    cin >> n;
    tr.resize(n+1); 
    lvl.resize(n+1,0);
    par.resize(n+1,vector<int>(20,-1)); 
    prefix.resize(n+1,0);
    for(int i=0;i<n-1;i++){
        int s,d;
        cin >> s >> d;
        tr[s].pb(d);
        tr[d].pb(s);
    }

    dfs(1,-1);

    int m;
    cin >> m;
    vector<pair<int,int>> a;
    a.resize(m);    
    for(int i=0;i<m;i++){
        int s,d;
        cin >> s >> d;
        a[i].ff = s;
        a[i].ss = d;
    }

    long long ans=0;
    for(int mask=0;mask<(1<<m);mask++){
        fill(prefix.begin(),prefix.end(),0);
        for(int j=0;j<m;j++){
            if((mask>>j)&1){
                int src = a[j].ff;
                int dest  = a[j].ss;
                int LCA = lca(src,dest);
                prefix[src]++;
                prefix[dest]++;
                prefix[LCA] -= 2;
            }
        }
        dfs1(1,0);
        int bits = __builtin_popcount(mask);
        int x = 0;
        for(int i=1;i<=n;i++){
            if(prefix[i])x++;
        }

        if(bits%2==0){
            ans += (1ll<<(n-x-1));
        }else{
            ans -= (1ll<<(n-x-1));
        }
    }
    cout << ans;

}

int main(){
    // #ifndef ONLINE_JUDGE
    // freopen("input00.txt","r",stdin);
    // #endif
    IOS;
    pre();
    ll T = 1;
    //cin>>T;
    while(T--){
        solve();   
    }
    return 0;
}