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>
using namespace std;

using PP = pair<int, int>;
int n, m;
vector<PP> G[50];
int p[50], q[50];
long ps[50][50];

void rec(int st, int from, int prev, long path) {
  ps[st][from] = path;
  for (PP e : G[from]) {
    int to = e.first;
    int id = e.second;
    if (to == prev) continue;
    path ^= 1l << id;
    rec(st, to, from, path);
    path ^= 1l << id;
  }
}

long sub(int s) {
  long cons = 0;
  for (int k = 0; k < m; ++k) {
    if ((s >> k) & 1) {
      cons |= ps[p[k]][q[k]];
    }
  }
  
  int cnt = __builtin_popcountll(cons);
  return 1l << (n - 1 - cnt);
}

int main() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    G[a].push_back(PP(b, i));
    G[b].push_back(PP(a, i));
  }
  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> p[i] >> q[i];
    --p[i]; --q[i];
  }
  
  for (int i = 0; i < n; ++i) {
    rec(i, i, -1, 0l);
  }
  
  long sum = 0;
  for (int s = 0; s < (1 << m); ++s) {
    int sgn = __builtin_popcountll(s) % 2 == 0 ? 1 : -1;
    sum += sgn * sub(s);
  }
  cout << sum << endl;
}