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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <algorithm>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

vector<P> g[55];
vector<int> dfs(int from, int to, int pre = -1){
  for (auto p : g[from]) {
    if (p.first == pre) continue;
    if (to == p.first) {
      return {p.second};
    }
    vector<int> ans = dfs(p.first,to,from);
    if (!ans.empty()) {
      ans.push_back(p.second);
      return ans;
    }
  }
  return {};
}

int main() {
  int n,m;
  cin >> n;
  rep(i,n-1){
    int a,b;
    cin >> a >> b;
    --a; --b;
    g[a].emplace_back(b,i);
    g[b].emplace_back(a,i);
  }

  cin >> m;
  vector<ll> eset(m);
  rep(i,m){
    int a,b;
    cin >> a >> b;
    --a; --b;
    vector<int> v = dfs(a,b);
    for (int e : v) eset[i] |= 1ll << e;
  }

  ll ans = 0;
  rep(i,1<<m){
    ll mask = 0;
    rep(j,m) if (i>>j&1) mask |= eset[j];
    
    int white = __builtin_popcountll(mask);
    ll cnt = 1ll << (n-1-white);
    if (__builtin_popcountll(i)%2==0) {
      ans += cnt;
    } else {
      ans -= cnt;
    }
  }
  cout << ans << endl;
}