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;
using ll = long long;
using ld = long double;
using pint = pair<int,  int>;
using pll = pair<ll, ll>;
using edge = pair<int, ll>;
using Graph = vector<vector<int>>; 
using WeightedGraph = vector<vector<edge>>;
const ll INF = 1LL << 60;
const int INF32 = 1 << 29;
const ll MOD = 1000000007;

WeightedGraph G;

bool rec(int u, int p, int v, bitset<64> &path){
  if(u==v)return true;
  for(auto e : G[u]){
    if(e.first==p)continue;
    if(rec(e.first, u, v, path)){
      path.set(e.second);
      return true;
    }
  }
  return false;
}

int main(){
  int n; cin >> n;
  G.resize(n);
  for(int i = 0; i < n-1; ++i){
    int u,v;cin >> u >> v;
    --u; --v;
    G[u].push_back({v,i});
    G[v].push_back({u,i});
  }
  
  int m; cin >> m;
  vector<int> us(m), vs(m);
  for(int i = 0; i < m; ++i){
    cin >> us[i] >> vs[i];
    --us[i]; --vs[i];
  }
  
  vector<bitset<64>> paths(m);
  for(int i = 0; i < m; ++i){
    rec(us[i], -1, vs[i], paths[i]);
  }
  
  ll ans = 0;
  for(ll i = 0; i < (1LL<<m); ++i){
    bitset<64> val(0), bit(i);
    for(int j = 0; j < m; ++j){
      if(bit[j]) val |= paths[j];
    }
    ll rem = n-1 - val.count();
    ll tmp = (1LL<<rem);
    if(bit.count()%2==0)ans+=tmp;
    else ans -= tmp;
  }
  cout << ans << endl;
}
