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;
typedef long long ll;
typedef pair<int,int> P;

vector<P>g[60];
vector<int>es;

bool dfs(int v, int tv, int p=-1){
  //aがbに到達した
  if(v==tv)return true;
  for(P u:g[v]){
    if(u.first==p)continue;
    //aからbまでの経路はtrue(esに経路が入る)
    if(dfs(u.first,tv,v)){
      es.push_back(u.second);
      return true;
    }
  }
  return false;
}

int main(){
  int n; cin>>n;
  rep(i,n-1){
    int a,b; cin>>a>>b;
    a--; b--;
    g[a].push_back(P(b,i));
    g[b].push_back(P(a,i));
  }
  int m; cin>>m;
  vector<ll>Eset(m);
  rep(i,m){
    int a,b; cin>>a>>b;
    a--; b--;
    es=vector<int>();
    dfs(a,b);
    //辺のidをlong longで管理(1ならその辺を含む)
    for(int bit:es){
      Eset[i]|=1LL<<bit;
    }
  }
  //包除原理
  ll ans=0;
  //i=0のとき　全体を表す
  rep(i,1<<m){
    ll mask=0;
    //条件の辺を白で塗る
    rep(j,m){
      if(i>>j&1)mask|=Eset[j];
    }
    int white=__builtin_popcountll(mask);
    //決定していない辺の個数はn-1-white個
    ll cnt=1LL<<(n-1-white);
    int pc=__builtin_popcount(i);
    if(pc%2==0)ans+=cnt;
    else ans-=cnt;
  }
  cout<<ans<<endl;
}