Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int N=55;
int n,m,a[N],b[N],fa[N],d[N],s[N];
long long ans;
vector<int>e[N];
void dfs(int u,int f)
{
	fa[u]=f,d[u]=d[f]+1;
	for(auto v:e[u])
		if(v!=f)
			dfs(v,u);
}
void cal(int x,int y)
{
	if(d[x]<d[y])
		swap(x,y);
	while(d[x]>d[y])
	{
		s[x]=0;
		x=fa[x];
	}
	while(x!=y)
	{
		s[x]=s[y]=0;
		x=fa[x],y=fa[y];
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		e[u].push_back(v);
		e[v].push_back(u);
	}
	dfs(1,0);
	scanf("%d",&m);
	for(int i=0;i<m;i++)
		scanf("%d%d",&a[i],&b[i]);
	for(int i=0;i<(1<<m);i++)
	{
		for(int j=2;j<=n;j++)
			s[j]=1;
		for(int j=0;j<m;j++)
			if(i&(1<<j))
				cal(a[j],b[j]);
		int t=0;
		for(int j=2;j<=n;j++)
			t+=s[j];
		ans+=(1ll<<t)*(__builtin_popcount(i)&1?-1:1);
	}
	printf("%lld\n",ans);
	return 0;
}