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;
typedef long long ll;
#define pb push_back
#define fi first
#define se second
typedef pair<ll,ll> P;
using VP = vector<P>; using VVP = vector<VP>;
using VI = vector<ll>; using VVI = vector<VI>; using VVVI = vector<VVI>;
const int inf=1e9+7;
const ll INF=1LL<<61;
const ll mod=1e9+7;


int n;
vector<int> E[101010];
VI ok(n+1);

void dfs(int cu, int pa = -1) {
  ok[cu]=1;
  for(int to:E[cu]) {
    if (to != pa) {

      dfs(to,cu);
      
    }
  }
}

int main(){
  int i,j;
  cin>>n;
  vector<P> w;
  for(i=0;i<n-1;i++){
    int a,b;
    cin>>a>>b;
    a--; b--;
    E[a].pb(b);
    E[b].pb(a);
    w.pb(P(a,b));
  }
  int m;
  cin>>m;
  int u[m],v[m];
  for(i=0;i<m;i++){
    cin>>u[i]>>v[i];
    u[i]--;
    v[i]--;
  }

  VI val(n,0);

  for(i=0;i<n-1;i++){
    ok=VI(n,0);
    dfs(w[i].fi,w[i].se);
    for(j=0;j<m;j++){
      if(ok[u[j]]+ok[v[j]]==1) val[i]+=(1<<j);
    }
  }

  VI dp((1<<m),0);
  dp[0]=1;
  for(i=0;i<n-1;i++){
    for(j=(1<<m)-1;j>=0;j--){
      dp[(j|val[i])]+=dp[j];
    }
  }
  cout<<dp[(1<<m)-1]<<endl;



  return 0;
}