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>
using namespace std;
#define _GLIBCXX_DEBUG
#define rep(i, n) for(int i=0; i<n; i++)
#define rep2(i, m, n) for(int i=int(m); i<n; i++)
#define ALL(a) a.begin(), a.end()
using ll = long long;
using ld = long double;
using P = pair<ll, ll>;
const ll INF = 100100100100;
const int MOD = 1000000007;

struct Edge{
  int id, to;
  Edge(int i, int t): id(i), to(t) {}
};
using Graph = vector<vector<Edge>>;

Graph G;
ll path;

bool dfs(int v, int p, int v2){
  if (v == v2) return true;
  for (auto e : G[v]) {
    if (e.to == p) continue;
    if (dfs(e.to, v, v2)) {
      path |= 1ll<<e.id;
      return true;
    }
  }
  return false;
}

int main(){
  int N; cin >> N;
  G = Graph(N);
  rep(i, N-1){
    int a, b; cin >> a >> b;
    a--; b--;
    G[a].push_back(Edge(i, b));
    G[b].push_back(Edge(i, a));
  }

  int M; cin >> M;
  vector<ll> paths(M);
  rep(i, M){
    int u, v; cin >> u >> v;
    u--, v--;
    path = 0;
    dfs(u, -1, v);
    paths[i] = path;
  }
  
  
  ll ans = 0;
  rep(i, 1ll<<M){
    ll eset = 0;
    rep(j, M) {
      if (i>>j & 1) {
        eset |= paths[j];
      }
    }
    ll num = 1ll<<(N-1 - __builtin_popcountll(eset));
    if(__builtin_popcountll(i) % 2 == 0) ans += num;
    else ans -= num;
  }
  cout << ans << endl;
  return 0;
}