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;
vector<pair<int,int> >v[53];
vector<int>path;
bool ispth[23][53];
long long n,m,dp[53][1048576],x[53],y[53];
void dfs(int x,int p,int t,int y){
	if(x==t)
		for(int i=0;i<path.size();i++)
			ispth[y][path[i]]=1;
	for(int i=0;i<v[x].size();i++)
		if(v[x][i].first!=p){
			path.push_back(v[x][i].second);
			dfs(v[x][i].first,x,t,y);
			path.pop_back();
		}
}
int main(){
	cin>>n;
	for(int i=1;i<n;i++){
		cin>>x[i]>>y[i];x[i]--;y[i]--;
		v[x[i]].push_back(make_pair(y[i],i));
		v[y[i]].push_back(make_pair(x[i],i));
	}cin>>m;
	for(int i=0;i<m;i++){
		int s,t;
		cin>>s>>t;
		s--;t--;
		dfs(s,-1,t,i);
	}
	dp[0][0]=1;
	for(int i=1;i<n;i++)
		for(int j=0;j<(1<<m);j++){
			dp[i][j]+=dp[i-1][j];
			int x=0;
			for(int k=0;k<m;k++)
				if(ispth[k][i]==1)
					x+=(1<<k);
			dp[i][j|x]+=dp[i-1][j];
		}
	cout<<dp[n-1][(1<<m)-1];
}