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>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;
struct edge{ll to,id;edge(ll to, ll id): to(to), id(id) {}};
vector<vector<edge> > v(50);
vector<ll> root;
bool dfs(int now,int p,int f){
	if(now==f)return true;
	for(edge i:v[now]){
		if(i.to==p)continue;
		if(dfs(i.to,now,f)){
			root.pb(i.id);
			return true;
		}
	}
	return false;
}
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n,m,a,b;
    cin>>n;
    for(ll i=1LL;i<n;i++){
    	cin>>a>>b;
    	a--;b--;
    	v[a].push_back({b,i-1});
    	v[b].push_back({a,i-1});
    }
    cin>>m;
    vector<ll> s(m,0LL);
    for(ll i=0LL;i<m;i++){
    	cin>>a>>b;
    	a--;b--;
    	root=vector<ll>();
    	bool niko=dfs(a,-1LL,b);
    	for(ll j:root){
    		s[i]|=1LL<<j;
    	}
    }
    ll ans=0,maki=(1LL<<m);
    for(ll bit=0;bit<maki;bit++){
    	ll mask=0;
    	for(ll i=0LL;i<m;i++){
    		if(bit>>i&1)mask|=s[i];
    	}
    	ll num=__builtin_popcountll(mask);
    	ll count=1LL<<(n-1-num);
    	if(__builtin_popcountll(bit)%2==0)ans += count;
    	else ans -= count;
    }
    cout << ans << endl;
    return 0;
}