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 ll=long long;
const ll MOD=(ll)1e9+7;
const ll INF=(ll)1e18+10;
const double PI=acos(-1);

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 (a>b) { a=b; return 1; } return 0; }
int dy[]={-1,1,0,0};
int dx[]={0,0,-1,1};


vector<int> tonari[55];
ll dp[1<<20][51];

int main() {
    int N; cin>>N;
    vector<vector<int>> next_path(N,vector<int>(N,-1));
    int a,b;
    for(int i=0;i<N-1;++i){
        cin>>a>>b; --a; --b;
        tonari[a].push_back(b);
        tonari[b].push_back(a);
        next_path[a][b]=i;
        next_path[b][a]=i;
    }

    int M; cin>>M;
    vector<int> path[21];

    for (int i = 0; i < M; ++i) {
        int u,v;
        cin>>u>>v; --u; --v;
        vector<int> prev(N,-1);
        queue<int> q;
        q.push(u);
        prev[u]=N;
        while(prev[v]==-1){
            int node=q.front();
            //cout<<node<<endl;
            q.pop();
            for(auto tsugi:tonari[node]) {
                if (prev[tsugi] >= 0) continue;
                prev[tsugi] = node;
                q.push(tsugi);
            }
        }
        //cout<<"ok"<<endl;
        int pos =v;
        while(prev[pos]!=N){
            path[i].push_back(next_path[pos][prev[pos]]);
            pos=prev[pos];
        }
        sort(path[i].begin(),path[i].end());
        /*
        cout<<i<<":";
        for(auto p:path[i]) cout<<p<<" ";
        cout<<endl;
        */
    }
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(int i=0;i<N;++i){
        int bits=0;
        for (int k = 0; k < M; ++k) {
            if(binary_search(path[k].begin(),path[k].end(),i)) bits|=(1<<k);
        }
        //cout<<i<<" "<<bits<<endl;
        for (int j = 0; j < (1 << M); ++j) {
            dp[j][i+1]+=dp[j][i];
            //int tmp=j;

            //printf("%d,%d : %d \n",j,i,tmp);
            dp[j|bits][i+1]+=dp[j][i];
        }
    }
    cout<<dp[(1<<M)-1][N]/2<<endl;

    return 0;
}