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 <stdio.h>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, m, u, v, ans = 0, cnt, now;
  scanf("%lld", &n);
  vector<ll> vec[n], num[n];
  for(i = 0; i < n - 1; ++i) {
    scanf("%lld%lld", &u, &v);
    vec[--u].push_back(--v);
    vec[v].push_back(u);
    num[u].push_back(i);
    num[v].push_back(i);
  }
  scanf("%lld", &m);
  vector<ll> a[m];
  ll b[n], c[n];
  for(i = 0; i < m; ++i) {
    scanf("%lld%lld", &u, &v);
    queue<ll> q;
    now = --u, --v;
    for(j = 0; j < n; ++j) b[j] = -1;
    b[now] = now;
    q.push(now);
    while(q.size()) {
      now = q.front(), q.pop();
      for(j = 0; j < vec[now].size(); ++j) if(b[vec[now][j]] == -1) b[vec[now][j]] = now, c[vec[now][j]] = num[now][j], q.push(vec[now][j]);
    }
    now = v;
    while(b[now] != now) a[i].push_back(c[now]), now = b[now];
  }
  bool flg[n];
  for(i = 0; i < (1 << m); ++i) {
    cnt = 0, now = 0;
    for(j = 0; j < n; ++j) flg[j] = true;
    for(j = 0; j < m; ++j) if((i >> j) & 1) {
      ++cnt;
      for(k = 0; k < a[j].size(); ++k) flg[a[j][k]] = false;
    }
    for(j = 0; j < n - 1; ++j) if(flg[j]) now++;
    ans += (1ll << now) * (cnt & 1 ? -1 : 1);
  }
  printf("%lld", ans);
  return 0;
}