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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;


vector<vector<P>> g(51);
vector<ll> es;
bool dfs(ll v, ll tv, ll p=-1){
  if (v==tv)return true;
  for(P e:g[v]){
    ll to=e.first;
    if(to==p) continue;
    ll id=e.second;
    if (dfs(to,tv,v)){
      es.push_back(id);
      return true;
    }
  }
  return false;
}
int main() {
  ll n, m;
  cin >> n;
  rep(i,n-1) {
    ll a, b;
    cin >> a >> b;
    --a; --b;
    g[a].emplace_back(make_pair(b, i));
    g[b].emplace_back(make_pair(a, i));
  }
  cin>>m;
  vector<ll> ebit(m);
  rep(i,m){
    ll a1,b1;
    cin>>a1>>b1;
    a1--;b1--;
    es=vector<ll>();
    dfs(a1,b1);
    for(auto enow:es){
      ebit[i]|=(1ll<<enow);
    }
  }
//  cout<<ebit[0]<<'k'<<endl;
  ll ans=0;
  for(ll i=0;i<(1ll<<m);i++){

    ll cnt=0;
    rep(j,m){
      if ((i>>j)&1){
        cnt|=ebit[j];
      //  cout<<cnt<<'c'<<j<<'j'<<ebit[j]<<endl;
      }
    }
    ll white = __builtin_popcountll(cnt);
    //cout<<white<<endl;
    ll now=(1ll<<(n-1-white));
     if (__builtin_popcount(i)%2==0){
       ans+=(1ll<<(n-1-white));
     }
     if (__builtin_popcount(i)%2==1){
       ans-=(1ll<<(n-1-white));
     }
  //   cout<<i<<'i'<<now<<endl;
  }
cout<<ans;
}
