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 int long long
#define pint pair<int,int>

int N;
vector<vector<int>> to;
int M;
vector<pint> P;

int sum = 0LL;

int num = 0LL; map<pint,int> mp;

vector<vector<int>>edges; //隣接行列

void dfs(int par,int pos){
  //0起点のdfsで、辺にナンバリングしていく
  if(par != -1LL ){
    //一番初めは記録しようがない
    mp[{par,pos}] = num; 
    mp[{pos,par}] = num; 
    //向きなしで記録する
    num++; //木だから、同じ辺を二回更新することはないはず
  }
 
  for(auto v : to[pos]){
    if(v == par){ continue;}
    
    dfs(pos,v);
  }
  
  return;
}

void ddfs(int par, int pos, int start,int val){
  
  int res = val;
  
  if( par != -1 ){ res |= (1LL << mp[{par,pos}]);}
  
  if( par != -1 ){ 
    //par == -1の時、pos == startだから、同じ点同士を結ぶ辺を記録することになり無意味
    

    edges[start][pos] |= res;
    edges[pos][start] |= res; //こちらも向きなしで記録する
    
  }
   
  for(auto u : to[pos]){
    if( u == par ){ continue;}
    
    ddfs(pos,u,start,res); //startは変化なし
  
  }
  
  return ;
}

  
signed main(void){ 
  
  cin >> N;
  to.resize(N);
  edges = vector<vector<int>> (N,vector<int>(N));
  
  for(int i = 1; i < N; i++){
    int a,b; cin >> a >> b; a--; b--;
    to[a].push_back(b); to[b].push_back(a);
  }
  
  dfs(-1LL,0LL);
  //ここまでOK
  
  for(int i = 0LL; i < N; i++){ 
    ddfs(-1LL,i,i,0LL);
  }
  
  //ここまでOK
  
  cin >> M; P.resize(M); //条件を列挙する
  
  for(int i = 0LL; i < M; i++){
    int u,v; cin >> u >> v; u--; v--;
    //uとvの間は少なくとも一つは黒で塗る必要がある
    P[i] = {u,v};
  }
   
  for(int bit = 1LL; bit < (1LL << M); bit++){
    //立ったbit全ての積を取って、包除原理を適用 i = 0,,,M-1はそれぞれの条件を使うか表す
    
    int value = 0LL;
    bool even = true;
    
    for(int i = 0; i < M; i++){
      if( (bit >> i) & 1LL ){  // P[i].firstとP[i].secondを結ぶ辺全てを記録
        even = !even;
        //i番目の条件に含まれている辺全てを追加
        value |= edges[P[i].first][P[i].second];
        
      }
    }
    //valueに含まれているbit全てを白で塗る必要がある
    
    int p = 0LL;
    
    for(int i = 0LL; i < N; i++){ //valueはNケタぐらいまではいく
      if( (value >> i) & 1LL ){ p++;}
    }
    
    int res = 0LL;
    
    if(!even){
      res += pow(2LL,N-1LL-p);
    }
    else{
      res -= pow(2LL,N-1LL-p);
    }
    
    sum += res;
    
  }
  
  cout << pow(2LL,N-1LL)-sum << endl;
  
  return 0;
}