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 long long long
using namespace std;
const int N=1e6,inf=1e9;
int parn[N+2],lv[N+2],n;
vector<int>adj[50+2];
map<pair<int,int>,int>mp;
long dp[52][(1<<20)+2];
void dfs(int node,int par,int l)
{
  parn[node]=par,lv[node]=l;
  for(auto x:adj[node])
  {
    if(x==par)continue;
    dfs(x,node,l+1);
  }
}
int pre[N+2];
void upd(int u,int v,int id)
{
  if(u>v)swap(u,v);
  pre[mp[{u,v}]]|=(1<<(id-1));
}
void lca(int u,int v,int id)
{
  if(lv[u]<lv[v])swap(u,v);
  while(lv[u]>lv[v])
  {
    upd(u,parn[u],id);
    u=parn[u];
  }
  while(u!=v)
  {
    upd(u,parn[u],id);u=parn[u];

    upd(v,parn[v],id);v=parn[v];
  }
}
int q;
long dfs(int i,int msk)
{
  if(i==n)return (msk==((1<<q)-1));
  if(dp[i][msk]!=-1)return dp[i][msk];
  return dp[i][msk]=dfs(i+1,msk)+dfs(i+1,msk|pre[i]);
}
int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  cin>>n;
  for(int i=1;i<n;i++)
  {
    int u,v;cin>>u>>v;if(u>v)swap(u,v);
    adj[u].push_back(v);adj[v].push_back(u);
    mp[{u,v}]=i;
  }
  dfs(1,-1,0);
  cin>>q;
  for(int i=1;i<=q;i++)
  {
    int u,v;cin>>u>>v;lca(u,v,i);
  }
  // for(int i=1;i<n;i++)
  //   cout<<pre[i]<<" ** ";
  // cout<<endl;
  memset(dp,-1,sizeof(dp));
  cout<<dfs(1,0)<<endl;
  return 0;
}