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;
typedef pair<int, int> ii;

const int N = 51;
vector<int> edges[N];
ll path[N];
ll info[N];
int n, m, a, b;

void dfs(int par, int u, ll id) {
  path[u] = id;
  for (auto v : edges[u]) {
    if (v == par) continue;
    dfs(u, v, id | (1ll << v));
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  dfs(0, 1, 0);

  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> a >> b;
    info[i] = path[a] ^ path[b]; // length of path
  }

  ll ans = 0;
  for (int i = 0; i < (1 << m); ++i) {
    ll mask = 0;
    int cnt = 0;
    for (int j = 0; j < m; ++j) {
      if (i & (1 << j)) {
        mask |= info[j];
        ++cnt;
      }
    }
    int fcnt =  n - 1 - __builtin_popcountll(mask);
    ans += pow(2, fcnt) * (cnt & 1 ? -1ll : 1ll);
  }

  cout << ans << endl;

  return 0;
}
