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:
/*********************
*  Author: xuziyuan  * 
*********************/
 
#include <bits/stdc++.h>
 
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <LL,LL>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair
 
using namespace std;
 
LL n,m,x,y,re[60],ans=0;
vector <pii> g[60];
vector <LL> pass;
bool ok;
 
void dfs(LL pos,LL to,LL no)
{
  if(pos==to)
  {
    ok=true;
    return;
  }
  rep(i,g[pos].size())
  {
    if(g[pos][i].fi==no) continue;
    pass.pb(g[pos][i].se);
    dfs(g[pos][i].fi,to,pos);
    if(ok) return;
    pass.pop_back();
  }
}
 
int main()
{
  cin>>n;
  rep(i,n-1)
  {
    cin>>x>>y;
    g[x].pb(mpr(y,i));
    g[y].pb(mpr(x,i));
  }
  cin>>m;
  rep(i,m)
  {
    cin>>x>>y;
    ok=false;
    dfs(x,y,0);
    rep(j,pass.size()) re[i]|=(LL)pow(2,pass[j]);
    pass.clear();
  }
  repn(i,(1<<m)-1)
  {
    LL cnt=0,now=0;
    rep(j,m)
    {
      if((i&(1<<j))==0) continue;
      now|=re[j];
      cnt++;
    }
    LL c=0;
    while(now>0)
    {
      c+=now%2;
      now/=2;
    }
    if(cnt&1) ans+=(LL)pow(2,n-1-c);
    else ans-=(LL)pow(2,n-1-c);
  }
  cout<<(LL)pow(2,n-1)-ans<<endl;
  return 0;
}