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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#define REP(i,n) for(int i=0;i<(n);++i)
using namespace std;
typedef long long LL;

int main() {
  int N, M;
  cin >> N;
  LL dist[50][50];
  REP(i, N) {
    REP(j, N) {
      dist[i][j] = 1LL << 60;
    }
    dist[i][i] = 0;
  }
  REP(i, N-1) {
    int A, B;
    cin >> A >> B;
    --A; --B;
    dist[A][B] = 1LL << i;
    dist[B][A] = 1LL << i;
  }
  REP(k, N) {
    REP(i, N) {
      REP(j, N) {
        LL ndist = dist[k][i] + dist[k][j];
        if(ndist < dist[i][j]) {
          dist[i][j] = ndist;
          dist[j][i] = ndist;
        }
      }
    }
  }
  cin >> M;
  vector<LL> routes(M);
  REP(i, M) {
    int U, V;
    cin >> U >> V;
    --U; --V;
    routes[i] = dist[U][V];
  }
  int lmax = 1 << M;
  LL ret = 0;
  REP(i, lmax) {
    int ti = i;
    LL mask = 0;
    REP(j, M) {
      if((ti>>j)&1) {
        mask |= routes[j];
      }
    }
    LL used = __builtin_popcountll(mask);
    if(__builtin_parity(i) == 0) {
      ret += 1LL << (N - 1 - used);
    } else {
      ret -= 1LL << (N - 1 - used);
    }
  }
  cout << ret << endl;
}