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 <map>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define all(a) (a).begin(), (a).end()

using ll = long long;
using P = std::pair<int, int>;

int N;
std::vector<int> G[100];
int M;
std::vector<int> es;
std::map<P, ll> e;
ll mask[100];

bool dfs( int v, int to, int p )
{
  if( v == to )
    return true;

  for( int u : G[v] )
  {
    if( u == p )
      continue;

    if( dfs( u, to, v ) )
    {
      es.emplace_back( e[P(v,u)] );

      return true;
    }
  }

  return false;
}

int main()
{
  scanf( "%d", &N );

  ll p = 0;
  rep( i, N-1 )
  {
    int a, b;
    scanf( "%d%d", &a, &b );
    --a; --b;

    G[a].emplace_back( b );
    G[b].emplace_back( a );

    e[P(a,b)] = p;
    e[P(b,a)] = p;

    ++p;
  }

  scanf( "%d", &M );

  rep( i, M )
  {
    int u, v;
    scanf( "%d%d", &u, &v );
    --u; --v;

    es.clear();
    dfs( u, v, -1 );

    for( int ve : es )
      mask[i] |= 1ll<<ve;
  }

  ll ans = 0;

  rep( i, 1ll<<M )
  {
    ll bit = 0;
    ll cnt = 0, cntI = 0;

    rep( j, M ) if( i >> j & 1 )
      bit |= mask[j], ++cntI;

    rep( j, N-1 ) if( bit >> j & 1 )
      ++cnt;

    ans += (cntI%2==0?1:-1)*(1ll<<(N-1-cnt));
  }

  printf( "%lld\n", ans );

  return 0;
}