Problem: 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.

Examples

Input

3
1 2
2 3
1
1 3


Output

3


Input

2
1 2
1
1 2


Output

1


Input

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5


Output

9


Input

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


Output

62

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i=0;i<(n);++i)
#define fi first
#define se second
#define INF 1000000009
#define lINF 1000000000000000009
typedef long long ll;

struct Edge{
  int to,id;
  Edge(int to,int id): to(to),id(id) {}
};
vector<Edge>node[59];
vector<int>r;
ll bits[29];

bool dfs(int now,int last,int g){
  if(now==g){
    return 1;
  }
  rep(i,node[now].size()){
    int next=node[now][i].to;
    int id=node[now][i].id;
    if(next==last)continue;
    if(dfs(next,now,g)){
      r.emplace_back(id);
      return 1;
    }
  }
  return 0;
}

int main(){
  int n,m;
  cin>>n;
  rep(i,n-1){
    int a,b;
    cin>>a>>b;
    a--;b--;
    node[a].emplace_back(b,i);
    node[b].emplace_back(a,i);
  }
  cin>>m;
  rep(i,m){
    int a,b;
    cin>>a>>b;
    a--;b--;
    r=vector<int>();
    dfs(a,-1,b);
    rep(j,r.size()){
      bits[i]|=1ll<<r[j];
    }
  }
  ll ans=0;
  rep(i,1<<m){
    ll mask=0;
    rep(j,m){
      if(i>>j&1)mask|=bits[j];
    }
    int bcnt=__builtin_popcountll(mask);
    ll sum=1;
    rep(i,n-1-bcnt){
      sum*=2;
    }
    if(__builtin_popcount(i)%2)sum=-sum;
    ans+=sum;
  }
  cout<<ans<<endl;
  return(0);
}