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;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e18;

ll n,m,x[55];
vector<pll> v[55];
ll dp[1<<20];

bool dfs1(ll a,ll b,ll c,ll d) {
  if(a==c) {
    return 1;
  }
  for(auto& i:v[a]) {
    if(i.first!=b&&dfs1(i.first,a,c,d)) {
      x[i.second]^=1<<d;
      return 1;
    }
  }
  return 0;
}

int main() {
  cin >> n;
  vec a(n),b(n);
  for(ll i=1;i<n;i++) {
    ll a,b;
    cin >> a >> b;
    v[a].emplace_back(b,i);
    v[b].emplace_back(a,i);
  }
  cin >> m;
  for(ll i=0;i<m;i++) {
    ll a,b;
    cin >> a >> b;
    dfs1(a,0,b,i);
  }
  dp[0]=1;
  for(ll i=1;i<n;i++) {
    for(ll j=(1<<m)-1;j>=0;j--) {
      dp[j|x[i]]+=dp[j];
    }
  }
  cout << dp[(1<<m)-1] << endl;
}