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;
typedef double dd;
const int N = 1e4+5;
const int maxn = 1e9+5;
const dd eps = 1e-6;
const ll mod = 1e9+7;

struct edge{
  int to,id;
  edge(int to,int id): to(to),id(id){}  
};

vector<edge> p[55];
vector<ll> s;

bool dfs(int u,int v,int fa){
  if(u == v) return true;
  for(auto e:p[u]){
    if(e.to == fa) continue;
    if(dfs(e.to,v,u)){
      s.push_back(e.id);
      return true;
    }
  }
  return false;
}

int n,m,u,v;
ll check[55];

int main(){
  // freopen("input.txt","r",stdin);
  ios::sync_with_stdio(false);
  cin>>n;
  for(int i = 1;i < n;i++){
    cin>>u>>v;
    p[u].emplace_back(v,i);
    p[v].emplace_back(u,i);
  }
  cin>>m;
  for(int i = 1;i <= m;i++){
    cin>>u>>v;
    ll mask = 0;
    s.clear();
    dfs(u,v,-1);
    for(auto now:s){
      check[i] |= 1ll<<(now-1);
    }
  }
  // ll ans = (1<<(n-1));
  ll ans = 0;
  for(int i = 0;i < (1<<m);i++){
    ll mask = 0,cnt = 0;
    for(int j = 1;j <= m;j++){
      if(i&(1<<(j-1))) mask |= check[j],cnt++;
    }
    ll white = __builtin_popcountll(mask);
    ll now = 1ll<<(n-1-white);
    if(cnt&1) ans -= now;
    else ans += now;
  }
  cout<<ans<<endl;
  return 0;
}