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 <cstdio>
#include <vector>
#include <utility>
#include <map>

/*
\cap_{u,v} \cup_{i\in (u,v) path} E_i

g(N) = |{E | {u,v\in M | \cup_{i\in (u,v) path} E_i} = N}|
G(N) = |{E | {u,v\in M | \cup_{i\in (u,v) path} E_i} \subseteq N}|
     = |{E | {u,v\in M-N | \cap_{i\in (u,v) path} !E_i}}|
*/

int n, m;
std::vector<int> E[50];

long long int M[20];
long long int A[1<<20];
std::map<std::pair<int, int>, int> edges;

long long int dfs(int u, int v, int p){
  if(u == v) return 0;
  for(int x: E[u]){
    if(x == p) continue;
    long long int b = dfs(x, v, u);
    if(b != -1) return (1LL << edges[{u, x}]) | b;
  }
  return -1;
}

int main(){
  scanf("%d", &n);
  for(int i = 0; i < n-1; i++){
    int a, b;
    scanf("%d%d", &a, &b);
    E[a-1].push_back(b-1);
    E[b-1].push_back(a-1);
    edges[{a-1,b-1}] = i;
    edges[{b-1,a-1}] = i;
  }
  scanf("%d", &m);
  for(int i = 0; i < m; i++){
    int a, b;
    scanf("%d%d", &a, &b);
    M[i] = dfs(a-1, b-1, -1);   
  }
  long long int ans = 1LL << (n-1);
  for(int i = 1; i < (1<<m); i++){
    A[i] = A[i&(i-1)] | M[__builtin_ctzll(i)];
    long long int x = 1LL << (n-1-__builtin_popcountll(A[i]));
    ans += (1-2*__builtin_parity(i))*x;
  }

  printf("%lld\n", ans);
  return 0;
}
