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)

/*********************
*  Author: xuziyuan  * 
*********************/
 
#include <bits/stdc++.h>
 
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <LL,LL>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair
 
using namespace std;
 
LL n,m,x,y,re[60],ans=0;
vector <pii> g[60];
bool vis[60];
 
bool dfs(LL pos,LL to,LL k)
{
  vis[pos]=true;
  if(pos==to) return true;
  rep(i,g[pos].size())
  {
    if(vis[g[pos][i].fi]) continue;
    if(dfs(g[pos][i].fi,to,k))
    {
      re[k]|=(LL)pow(2,g[pos][i].se);
      return true;
    }
  }
  return false;
}
 
int main()
{
  cin>>n;
  rep(i,n-1)
  {
    cin>>x>>y;
    g[x].pb(mpr(y,i+1));
    g[y].pb(mpr(x,i+1));
  }
  cin>>m;
  rep(i,m)
  {
    cin>>x>>y;
    memset(vis,0,sizeof(vis));
    dfs(x,y,i);
  }
  repn(i,(1<<m)-1)
  {
    LL num=__builtin_popcount(i),now=0;
    rep(j,m)
    {
      if((i&(1<<j))==0) continue;
      now|=re[j];
    }
    LL res=(LL)pow(2,n-1-__builtin_popcount(now));
    if(num%2==1) ans+=res;
    else ans-=res;
  }
  cout<<(LL)pow(2,n-1)-ans<<endl;
  return 0;
}