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 N=205;
int n,m,dep[N],fa[N][10],c[N];
int v[N],nxt[N],head[N],a[N][2];
long long ans,res;

void dfs(int x)
{
	for(int i=1;i<=5;i++) fa[x][i]=fa[fa[x][i-1]][i-1];
	for(int i=head[x];i>0;i=nxt[i])
	{
		if(v[i]==fa[x][0]) continue;
		fa[v[i]][0]=x,dep[v[i]]=dep[x]+1,dfs(v[i]);
	}
}

void dfs2(int x)
{
//	cout<<x<<' '<<c[x]<<endl;
	for(int i=head[x];i>0;i=nxt[i])
	{
		if(v[i]==fa[x][0]) continue;
		dfs2(v[i]),c[x]+=c[v[i]];
	}
	if(c[x]) res++;
}

int lca(int x,int y)
{
	if(dep[x]<dep[y]) swap(x,y);
	int c=dep[x]-dep[y];
	for(int i=0;i<=5;i++) if(c&(1<<i)) x=fa[x][i];
	for(int i=5;i>=0;i--) if(fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i];
	return x==y?x:fa[x][0];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&v[i],&v[i+n]);
		nxt[i]=head[v[i+n]],head[v[i+n]]=i;
		nxt[i+n]=head[v[i]],head[v[i]]=i+n;
	}
	dfs(1),scanf("%d",&m);
	for(int i=1;i<=m;i++) scanf("%d%d",&a[i][0],&a[i][1]);
	for(int i=1;i<(1<<m);i++)
	{
		int cnt=0;
		for(int j=1;j<=n;j++) c[j]=0;
		for(int j=1;j<=m;j++)
		{
			if((i&(1<<j-1))==0) continue;
			cnt++;
			c[a[j][0]]++,c[a[j][1]]++;
			c[lca(a[j][0],a[j][1])]-=2;
		}
		res=0,dfs2(1);
//		cout<<res<<endl;
		if(cnt%2) ans+=(1ll<<n-1-res);
		else ans-=(1ll<<n-1-res);
	}
	printf("%lld\n",(1ll<<n-1)-ans);
}