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;
typedef long long ll;
const ll INF = 100000;
struct edge {int to, label;};
//隣接リスト
vector<edge> G[50];

bool dfs(int v, int p, int target, ll &path) {
  if (v == target) {
    return true;
  }
  for (int i; i < G[v].size(); i++) {
    edge e = G[v][i];
    if (e.to == p) continue;
    if (dfs(e.to, v, target, path)) {
      path |= (1LL << e.label);
      return true;
    }
  }
  return false;
}

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({b, i});
    G[b].push_back({a, i});
  }
  
  //path[i] はi番目のpathに含まれる辺たちを記録
  int M;
  cin >> M;
  
  ll path[M];
  for (int i = 0; i < M; i++) {
    path[i] = 0;
  }
  
  for (int i = 0; i < M; i++) {
    int u, v; cin >> u >> v; u--; v--;
    dfs(u, -1, v, path[i]);
  }
  
  //共通部分はor演算で求まる
  ll bits;
  ll ans = 0;
  
  for (bits = 0; bits < (1LL << M); bits++) {
    ll val = 0;
    for (int i = 0; i < M; i++) {
      if (bits & (1LL << i)) {val |= path[i];}
    }
    int c = __builtin_popcountl(val);
    int b = __builtin_popcountl(bits);
    if (b % 2 == 1) { ans -= 1LL<<(N-1-c); }
    else { ans += 1LL<<(N-1-c); }
  }
  cout << ans << endl;
}