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 rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;
struct edge
{
  ll to;
  ll cost;
  edge() {}
  edge(ll to, ll cost) : to(to), cost(cost){};
};
int main()
{

ll n;
cin >> n;
vector<vector<edge>> g(n);
rep(i, n-1)
{
  ll a, b;
  cin >> a >> b ;
  g[a - 1].emplace_back(b - 1, i);
  g[b - 1].emplace_back(a - 1, i); //無向辺
}
ll m ;
cin >> m ;
vector<ll> v;
rep(i,m){
  ll a, b;
  cin >> a >> b;
  a--;b--;
  vector<pair<ll,ll>> par(n);
  vector<bool> see(n);
  queue<ll> q;
  par[a]={-1,-1};
  see[a]=1;
  q.push(a);
  while(!q.empty()){
    auto p=q.front();
    if(p==b){
      ll vv=0;
      while(par[p].first!=-1){
        vv+=(1LL<<par[p].second);
        p=par[p].first;
      }
      v.push_back(vv);
      break;
    }
    see[p]=1;
    q.pop();
    for(auto r:g[p]){
      if(see[r.to]!=0)continue;
      par[r.to]={p,r.cost};
      q.push(r.to);
    }
  }
}
rep(i,1LL<<m){
  ll k=__builtin_popcountll(i);
  ll a=0;
  rep(j,m){
    if(i>>j & 1LL){
      a|=v[j];
    }
  }
  if(k%2)ans-=1LL<<(n-1-__builtin_popcountll(a));
  else ans+=1LL<<(n-1-__builtin_popcountll(a));
}
cout << ans << endl;
return 0;
}