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>
#define ll long long
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define eb emplace_back
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 233;
vector<int> g[maxn];
int f[maxn],u[maxn],v[maxn],d[maxn];
ll p[100];
bool vis[100];
void dfs(int x,int fa){
	f[x] = fa;
	for(int y:g[x]){
		if(y==fa)continue;
		d[y] = d[x]+1;
		dfs(y,x); 
	}
}
void lca(int x,int y){
	if(d[x]<d[y])swap(x,y);
//	cout<<"!!!!!!!!"<<x<<" "<<y<<endl;
	while(d[x]>d[y]){
		vis[x] = 1;
		x = f[x];
	}
	if(x==y)return ;
	while(x!=y){
		vis[x] = 1;
		x = f[x];
		vis[y] = 1;
		y = f[y];
	}
}
int main()
{
	int n;cin>>n;
	for(int i=1;i<n;i++){
		int x,y;scanf("%d%d",&x,&y);
		g[x].pb(y);
		g[y].pb(x);
	}
	d[1] = 1;
	dfs(1,0);
	int m;cin>>m;
	for(int i=0;i<m;i++){
		scanf("%d%d",&u[i],&v[i]);
	}
	p[0] = 1;
	for(int i=1;i<=50;i++)p[i] = p[i-1]*2;
	ll ans = 0;
	for(int i=0;i<(1<<m);i++){
		memset(vis,0,sizeof(vis));
		int op = 1;
		for(int j=0;j<m;j++)
			if(i&(1<<j)){
				op = -op;
				lca(u[j],v[j]);
			}
		int cnt = 0;
		for(int i=2;i<=n;i++)
			if(!vis[i])cnt++;
		ans += op*p[cnt];
//		cout<<"!"<<i<<" "<<cnt<<endl;
	}
	cout << ans << endl;
	return 0;
}