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>
#define rep(i,n) for (ll i = 0; i < n; ++i)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
const int INF=1001001001;

vector<ll> e;
vector<P> g[51];

bool dfs(ll t,ll p, ll l){
  if (t==l)return true;
  for(P v:g[t]){
    if (v.first==p)continue;


    if(dfs(v.first,t,l)){
      e.push_back(v.second);
    //  cout<<v.second<<"v"<<endl;
      return true;
    }
  }
  return false;
}

int main() {
  ll n;
  cin>>n;
  rep(i,n-1){
    ll a1,b1;
    cin>>a1>>b1;
    a1--;b1--;
    g[a1].push_back(make_pair(b1,i));
    g[b1].push_back(make_pair(a1,i));
  }
  ll m;
  cin>>m;
  vector<ll> edge(m);
  rep(i,m){
    ll u1,v1;
    cin>>u1>>v1;
    u1--;v1--;
    e=vector<ll>();
    //cout<<i<<"dfs"<<endl;
    dfs(u1,-1,v1);
    for(ll j:e){
      edge[i]|=(1ll<<j);
    }
  }
  ll ans=0;
  rep(i,1ll<<m){
    ll now=0;
    rep(j,m){
      if ((i>>j)&1){
        now|=edge[j];
      }
    }
    ll c=__builtin_popcountll(now);
    ll c1=__builtin_popcountll(i);
    //cout<<i<<'i'<<c<<"c"<<endl;
    if (c1%2==0)ans+=(1ll<<(n-1-c));
    else ans-=(1ll<<(n-1-c));
  }
  cout<<ans;
}