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 <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define endl '\n'
using ll=long long;
 
const int INF=1e9+5;
const ll lim=(ll)3e16;
const int N=3e3+5;
const int mod=1e9+7;
const ll oo=1e18+5;

vector<pair<int,int>> v[55];
vector<int> path,found;

void add_self(int& a,int b){
	a+=b;
	if(a>=mod) a-=mod;
}

void dfs(int a,int goal,int par){
	if(a==goal){
		found=path;
		return;
	}
	for(pair<int,int> x:v[a]){
		int b=x.first;
		if(b==par) continue;
		path.push_back(x.second);
		dfs(b,goal,a);
		path.pop_back();
	}
}

int main(){
	ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n-1;++i){
		int x,y;
		cin>>x>>y;
		v[x].push_back(mp(y,i));
		v[y].push_back(mp(x,i));
	}
	int m;
	cin>>m;
	vector<pair<int,int>> paths;
	vector<int> containMe(n-1 );
	for(int i=0;i<m;++i){
		int a,b;
		cin>>a>>b;
		found={-1};
		dfs(a,b,-1);
		for(int x:found){
			containMe[x]|=(1<<i);
		} 
	}
	vector<ll> dp(1<<m);
	dp[0]=1;
	for(int i=0;i<n-1;++i){
		// cout<<containMe[i]<<endl;
		vector<ll> new_dp(1<<m);
		for(int mask=0;mask<(1<<m);++mask){
			new_dp[mask]+=dp[mask];
			new_dp[mask|containMe[i]]+=dp[mask];
		}
		dp=new_dp;
	}
	cout<<dp[(1<<m)-1];
	return 0;
}
        
