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;
vector<vector<int>> g;
vector<int> par, mask;
map<pair<int, int>, int> e;
vector<pair<int, int>> p;
vector<vector<long long>> dp;
int n, m;

#ifdef LOCAL 
#include "/home/gokul/Codes/lib/prettyprint.hpp"
#else
#define debug(...) 2
#endif

void dfs(int x, int pr) {
  par[x] = pr;
  for (auto i : g[x]) 
    if (i != pr)
      dfs(i, x);  
}

long long solve(int x, int mk) {
  debug(x, mk);
  if (x == n-1) return (__builtin_popcount(mk) == m);
  long long &ret = dp[x][mk];
  if (ret != -1)  return ret;
  ret = (solve(x + 1, mk|mask[x]) + solve(x + 1, mk));
  return ret;
}

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);

  cin >> n;
  g.resize(n + 5);
  mask.resize(n + 5);
  par.resize(n + 5);

  for (int i = 0; i < n-1; ++i) {
    int u, v;
    cin >> u >> v;

    g[u].push_back(v);
    g[v].push_back(u);
    e[{u, v}] = e[{v, u}] = i;
    p.push_back({u, v});  
  }

  cin >> m;
  dp.assign(n + 5, vector<long long> ((1<<m), -1));
  
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    dfs(u, u);
    while (v != par[v]) {
      int id = e[{v, par[v]}];
      mask[id] |= (1<<i);
      v = par[v];
    }
  }

  debug(mask);
  cout << solve(0, 0) << '\n';
}

