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 <iostream>
#include <bitset>
#include <vector>
using namespace std;

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

bitset<55> ans[22];

vector<P> g[55];
bool ok;
ll fac[55];
ll an;

void dfs(int m, int s, int prev, int t) {
 
  if(s == t) {
    ok = true;
    return;
  }

  for(int i = 0; i < g[s].size(); i++) {
    int u = g[s][i].first; int id = g[s][i].second;
    if(u == prev) continue;
    dfs(m,u,s,t);
    if(ok) {
      ans[m][id] = true;
      return;
    }
  }

  return;
}
    
				       
  
  

int main() {
  int n;
  cin >> n;
  for(int i = 0; i < n - 1; i++) {
    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;
  for(int i = 0; i < m; i++) {
    int x,y;
    cin >> x >> y;
    x--; y--;
    ok = false;
    dfs(i,x,-1,y);
  }
  fac[0] = 1;
  
  for(int i = 1; i < 55; i++) fac[i] = fac[i-1] * 2;
  
 

  for(int i = 1; i < (1 << m); i++) {
    int cnt = 0;
    bitset<55> tmp;
    for(int j = 0; j < 20; j++) {
      int s = (1 << j);
      if(s & i) {
	tmp |= ans[j];
	cnt++;
      }

    }
  
    if(cnt & 1) {
      an += fac[n-1-tmp.count()];
    
    }
    else an -= fac[n-1-tmp.count()];
    //   cout << cnt <<"  "<< tmp <<"   "<<n-1-tmp.count()<< endl;
  }

  cout << fac[n-1] - an << endl;
  
}
  
