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
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 55 + 10;
#define fi first
#define se second
#define pb push_back
int n,m;
struct uzi{
  int s,t,i;
}p[N];
vector<int>v[N];
vector<int>g[N][N];
int b[51][51],pre[N];
void dfs(int x,int y){
  pre[x]=y;
  for(auto k:v[x])if(k!=y){
      dfs(k,x);
    }
}
bitset<50>li[N];
int main() {
  ios::sync_with_stdio(false);
  cin>>n;
  for(int i=1;i<n;i++){
    int s,t;
    cin>>s>>t;
    v[s].pb(t);
    v[t].pb(s);
    b[s][t]=b[t][s]=i;
  }
  for(int i=1;i<=n;i++){
    for(int j=1;j<=n;j++)pre[j]=0;
    dfs(i,0);
    for(int j=1;j<=n;j++){
      int now=pre[j];
      while(now){
        g[i][j].pb(now);
        now=pre[now];
      }
    }
  }
  cin>>m;
  LL ans=1ll<<n-1;
  // cout<<ans<<endl;
  for(int i=0;i<m;i++){
    cin>>p[i].s>>p[i].t;
    int S=p[i].s,t=p[i].t;
    for(auto z:g[S][t]){
      li[i].set(b[t][z]-1);
      //if(i==2)cout<<t<<' '<<z<<endl;
      t=z;
    }
  }
  for(int i=1;i<1<<m;i++){
    vector<int>x;
    for(int j=0;j<m;j++){
      if(1<<j&i) {
        x.pb(j);
      }
    }
    int y=__builtin_popcount(i)&1?-1:1;
    bitset<50>ggx;ggx.reset();
    for(auto k:x){
      ggx|=li[k];
    }
    ans+=y*(1ll<<(n-1-ggx.count()));
  }
  cout<<ans<<'\n';
  return 0;
}
