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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll MAX_V = 60;
vector<pair<ll, ll> > G[MAX_V];
// a -> bにいくときに通る辺のbits
ll E[MAX_V][MAX_V];
// 制約iで通る辺のbits
ll EM[30];

void dfs(ll v, ll parent, ll root, ll mask) {
  E[root][v] = mask;
  for(auto kv : G[v]) {
    if(kv.first == parent)continue;
    dfs(kv.first, v, root, mask | kv.second);
  }
}

int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);

  ll n;cin>>n;
  for(ll i=0;i<n-1;i++){
    ll a,b;cin>>a>>b;a--,b--;
    G[a].push_back({b,1ll<<i});G[b].push_back({a,1ll<<i});
  }
  for(ll i=0;i<n;i++) dfs(i, -1, i, 0);
  ll m;cin>>m;
  for(ll i=0;i<m;i++){
    ll u,v;cin>>u>>v;u--,v--;
    EM[i] = E[u][v];
  }
  // 包除原理
  ll ans=0;
  for(ll i=0;i<(1ll<<m);i++){
    ll whites = 0;
    for(ll j=0;j<m;j++){
      if(i & (1ll << j)) whites |= EM[j];
    }
    ll cnt = pow(2, n - 1 - __builtin_popcountll(whites));
    if (__builtin_popcountll(i) % 2 == 0) {
      ans += cnt;
    } else {
      ans -= cnt;
    }
    // cout<<cnt<<endl;
  }
  cout<<ans<<endl;

  return 0;
}