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;

using i64 = long long;

vector<bitset<50>> use_edge;
vector<vector<pair<int, int>>> edge;

bool dfs(int v, int i, int n, int p = -1) {
  if(n == v) return true;
  
  for(auto &e : edge[n]) {
    if(e.first == p) continue;
    
    bool res = dfs(v, i, e.first, n);
    
    if(res) {
      use_edge[i].set(e.second);
      return true;
    }
  }
  return false;
}

int main() {
  int n;
  cin >> n;
  edge = vector<vector<pair<int, int>>>(n);
  for(int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    edge[a - 1].push_back({ b - 1, i });
    edge[b - 1].push_back({ a - 1, i });
  }
  int m;
  cin >> m;
  use_edge = vector<bitset<50>>(m, bitset<50>(0));
  for(int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    dfs(v - 1, i, u - 1);
  }

  i64 cnt = 0;
  for(int bit = 1; bit < (1 << m); ++bit) {
    int popcnt = 0;
    bitset<50> b(0);
    
    for(int i = 0; i < m; ++i) {
      if(bit & (1 << i)) {
        b |= use_edge[i];
        popcnt++;
      }
    }
    
    int e = (n - 1) - b.count();
    
    if(popcnt & 1 == 1) cnt += (1ll << e);
    else cnt -= (1ll << e);
  }

  i64 ans = (1ll << (n - 1)) - cnt;
  cout << ans << endl;
}
