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<iostream>
#include<cstdio>
using namespace std;
int n,ccnt,sum,m;
int fir[1010],to[2020],nxt[2020],fa[1010],dep[1010];
int Lca[101][101],u[101],v[101],cnt[101];
long long f2[1010];
void add(int x,int y)
{
	nxt[++ccnt]=fir[x];fir[x]=ccnt;to[ccnt]=y;
}
void dfs(int x)
{
	dep[x]=dep[fa[x]]+1;
	for(int i=fir[x];i;i=nxt[i])
		if(to[i]!=fa[x])
		{
			fa[to[i]]=x;
			dfs(to[i]);
		}
}
void calc(int x)
{
	for(int i=fir[x];i;i=nxt[i])
		if(to[i]!=fa[x])
		{
			calc(to[i]);
			cnt[x]+=cnt[to[i]];
		}
	if(cnt[x]>0) sum++;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	dfs(1);
	for(int i=1;i<=n;i++)
		for(int j=i;j<=n;j++)
		{
			int x=i,y=j;
			if(dep[x]<dep[y]) swap(x,y);
			while(dep[x]>dep[y]) x=fa[x];
			while(x!=y) x=fa[x],y=fa[y];
			Lca[i][j]=Lca[j][i]=x;
		}
	scanf("%d",&m);
	for(int i=1;i<=m;i++) scanf("%d%d",&u[i],&v[i]);
	f2[0]=1;
	for(int i=1;i<=n;i++) f2[i]=f2[i-1]<<1;
	long long ans=0;
	for(int i=0;i<(1<<m);i++)
	{
		long long fl=1;
		for(int j=1;j<=m;j++)
			if((i>>(j-1))&1)
				cnt[u[j]]++,cnt[v[j]]++,cnt[Lca[u[j]][v[j]]]-=2,fl=-fl;
		sum=0;
		calc(1);
		ans+=fl*f2[n-1-sum];
		for(int j=1;j<=n;j++) cnt[j]=0;
	}
	printf("%lld\n",ans);
	return 0;
}