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;
#define ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<n;i++)
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl; 
#define INF 1000000000
#define mod 1000000007
using ll=long long;
const ll LINF = 1001002003004005006ll;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

struct edge{
    int id,to;
    edge(int to,int id) : to(to),id(id){}
};

int n,m;
vector<vector<edge>> g;
vector<int> ids;
bool dfs(int pre,int now,int gl){
    if(now==gl){
        return true;
    }
    for(auto x:g[now])if(x.to!=pre){
        if(dfs(now,x.to,gl)){
            ids.push_back(x.id);
            return true;
        }
    }
    return false;
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(0);

    cin>>n;
    g.resize(n);
    rep(i,n-1){
        int a,b;cin>>a>>b;a--,b--;
        g[a].push_back(edge(b,i));
        g[b].push_back(edge(a,i));
    }
    cin>>m;
    vector<ll> es(m);
    rep(i,m){
        int u,v;cin>>u>>v;u--,v--;
        ids.clear();
        dfs(-1,u,v);

        ll b=0;
        for(auto x:ids){
            b|=(1ll<<x);
        }
        es[i]=b;
    }

    ll ans=0;
    rep(bit,(1<<m)){
        ll mask=0;
        rep(i,m){
            if((bit>>i)&1) mask|=es[i];
        }
        ll tmp=(1ll<<(n-1-__builtin_popcountll(mask)));
        ans+=__builtin_parity(bit)?-tmp:tmp;
    }
    cout<<ans<<endl;
    return 0;
}