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;
typedef pair<int,int> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=1e9+7;

int main() {
  int n;
  cin>>n;
  vector<int> path[n];
  rep(i,0,n-1){
    int a,b;
    cin>>a>>b;
    path[a-1].push_back(b-1);
    path[b-1].push_back(a-1);
  }
  int m;
  cin>>m;
  vector<P> abc[m];
  rep(i,0,m){
    int u,v;
    cin>>u>>v;
    u--;
    v--;
    int dp[n];
    rep(j,0,n) dp[j]=810;
    dp[u]=0;
    queue<int> q;
    q.push(u);
    while(q.size()){
      int r=q.front();
      q.pop();
      for(int x:path[r]){
        if(dp[x]==810){
          dp[x]=dp[r]+1;
          q.push(x);
        }
      }
    }
    int cur=v;
    while(cur!=u){
      for(int x:path[cur]){
        if(dp[cur]==dp[x]+1){
          P p=P(cur,x);
          if(p.first>p.second) swap(p.first,p.second);
          abc[i].push_back(p);
          cur=x;
          break;
        }
      }
    }
  }
  ll power[n];
  power[0]=1;
  rep(i,0,n-1) power[i+1]=power[i]*2;
  ll ans=0;
  rep(i,0,1<<m){
    set<P> st;
    int cnt=0;
    rep(j,0,m){
      if(i>>j & 1){
        cnt++;
        for(P pp:abc[j]){
          st.insert(pp);
        }
      }
    }
    if(cnt%2==0){
      ans+=power[n-1-st.size()];
    }
    else{
      ans-=power[n-1-st.size()];
    }
  }
  cout<<ans<<endl;
}