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;
const int MAXN=110;
int tot,head[MAXN];
struct Edge {int to,net;}E[MAXN<<1];
void addedge(int x,int y)
{
	E[++tot].to=y;E[tot].net=head[x];head[x]=tot;
	E[++tot].to=x;E[tot].net=head[y];head[y]=tot;
}
long long Ans,Pow[MAXN];
int depth[MAXN],fa[MAXN];
int N,M,A[MAXN],B[MAXN];
int Cnt[MAXN];
void dfs(int x)
{
	for (int i=head[x];i;i=E[i].net)
		if (E[i].to!=fa[x])
		{
			fa[E[i].to]=x;
			depth[E[i].to]=depth[x]+1;
			dfs(E[i].to);
		}
}
void solve(int x,int y,int v)
{
	while (x!=y)
	{
		if (depth[x]<depth[y]) swap(x,y);
		Cnt[x]+=v;
		x=fa[x];
	}
}
void dfs2(int x,int cnt)
{
	if (x==M+1)
	{
		int sum=0;
		for (int i=2;i<=N;++i)
			if (!Cnt[i])
				sum++;
		if (cnt&1) Ans-=Pow[sum];
		else Ans+=Pow[sum];
		return ;
	}
	solve(A[x],B[x],1);
	dfs2(x+1,cnt+1);
	solve(A[x],B[x],-1);
	dfs2(x+1,cnt);
}
int main()
{
	Pow[0]=1;
	for (int i=1;i<50;++i) Pow[i]=Pow[i-1]*2;
	cin>>N;
	for (int i=1;i<N;++i)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		addedge(a,b);
	}
	dfs(1);
	cin>>M;
	for (int i=1;i<=M;++i) cin>>A[i]>>B[i];
	dfs2(1,0);
	cout<<Ans<<endl;
	return 0;
}