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
import java.math.*;
import java.util.*;
import java.util.function.*;
 
public class Main {
  private static long path(int[][] edge, int parent_node, int src_node, int target_node) {
    int[] dist_edges = edge[src_node];
    for (int dist_node = 1; dist_node < dist_edges.length; dist_node++) {
      int edge_no = dist_edges[dist_node];
      if (edge_no == 0) continue;
      if (dist_node == parent_node) continue;
      if (dist_node == target_node) return 1L << edge_no - 1;
      long path = path(edge, src_node, dist_node, target_node);
      if (path == 0) continue;
      return  (1L << edge_no - 1) | path;
    }
    return 0L;
  }

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int[] a = new int[N];
    int[] b = new int[N];
    for (int i = 1; i < N; i++) {
      a[i] = sc.nextInt();
      b[i] = sc.nextInt();
    }
    int M = sc.nextInt();
    int[] u = new int[M];
    int[] v = new int[M];
    for (int i = 0; i < M; i++) {
      u[i] = sc.nextInt();
      v[i] = sc.nextInt();
    }
    sc.close();

    int[][] edge = new int[N + 1][N + 1];
    for (int i = 1; i < N; i++) {
      edge[a[i]][b[i]] = i;
      edge[b[i]][a[i]] = i;
    }

    long[] path = new long[M];
    for (int i = 0; i < M; i++)
      path[i] = path(edge, 0, u[i], v[i]);

    long ans = 0;
    long mmax = 1L << M;
    for (long i = 0; i < mmax; i++) {
      long p = 0;
      long m = i;
      for (int j = 0; m != 0 && j < M; j++) {
        if ((m & 1) == 1) p |= path[j];
        m >>>= 1;
      }
      long ng_case = 1L << (N - 1 - Long.bitCount(p));
      if ((Long.bitCount(i) & 1) == 1) {
        ans -= ng_case;
      } else {
        ans += ng_case;
      }
    }
    System.out.println(ans);
  }
}