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
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define int long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int dp[55][(1<<20)],n,m;
V<int> v[55],g[55];
int fa[55],de[55];
int er[25];
void dfs(int p){
	for(int i=0;i<v[p].size();i++){
		if(fa[p]!=v[p][i]){
			fa[v[p][i]]=p;
			de[v[p][i]]=de[p]+1;
			dfs(v[p][i]);
		}
	}
}
void get(int l,int r,int s){
	if(de[l]>de[r])swap(l,r);
	while(de[r]>de[l]){
		g[r].PB(s);r=fa[r];
	}
	while(l!=r){
		g[l].PB(s);l=fa[l];
		g[r].PB(s);r=fa[r];
	}
} 
signed main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	int x,y;
	cin>>n;
	FOR(i,2,n){
		cin>>x>>y;v[x].PB(y);v[y].PB(x);
	}
	fa[1]=-1;de[1]=1;
	dfs(1);
	cin>>m;
	er[0]=1;
	FOR(i,1,m)er[i]=er[i-1]*2;
	FOR(i,1,m){
		cin>>x>>y;get(x,y,i);
	}
//	FOR(i,1,n){
//		for(int j=0;j<g[i].size();j++)cout<<g[i][j]<<' ';
//		cout<<'\n';
//	}
	dp[1][0]=1;
	FOR(i,2,n){
		FOR(j,0,er[m]-1){
			dp[i][j]=dp[i][j]+dp[i-1][j];
			int t=j;
			for(int k=0;k<g[i].size();k++)t=t|(er[g[i][k]-1]);
			dp[i][t]=dp[i][t]+dp[i-1][j];
		}
	}
	cout<<dp[n][er[m]-1];
	RE 0;
}


