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<iostream>
#include<vector>

struct edge{
  int to;
  int ind;
};

std::vector<std::vector<edge>> G(55);
std::vector<long long> paths(55);
int N,M,U[22],V[22];

long long dfs(int v,int tar,int par,long long path){
  if(v==tar){
    return path;
  }
  for(edge e:G[v]){
    if(e.to==par) continue;
    long long res=dfs(e.to,tar,v,path+(1LL<<e.ind));
    if(res!=-1) return res;
  }
  return -1;
}

void solve(){
  std::cin>>N;
  for(int i=0;i<N-1;i++){
    int A,B;
    std::cin>>A>>B;
    A--,B--;
    G[A].push_back({B,i});
    G[B].push_back({A,i});
  }
  std::cin>>M;
  for(int i=0;i<M;i++){
    std::cin>>U[i]>>V[i];
    U[i]--,V[i]--;
    paths[i]=dfs(U[i],V[i],-1,0);
  }
  
  long long ans=0;
  for(int bit=1;bit<(1<<M);bit++){
    long long orpath=0,bit_popcount=0;
    for(int i=0;i<M;i++){
      if((bit>>i)&1){
        orpath|=paths[i];
        bit_popcount++;
      }
    }
    int path_popcount=0;
    for(int i=0;i<N-1;i++){
      if((orpath>>i)&1) path_popcount++;
    }
    if(bit_popcount%2) ans+=1LL<<(N-1-path_popcount);
    else ans-=1LL<<(N-1-path_popcount);
  }
  std::cout<<(1LL<<(N-1))-ans<<std::endl;
  return;
}

int main(){
  solve();
  return 0;
}