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 MAXN 101
using namespace std;
int n,m,head[MAXN],tot,pre[MAXN],bel[MAXN],deep[MAXN];
long long ans;
vector<int> P[MAXN];
bool mark[52];
struct node{
	int ed,num,last;
}G[MAXN<<2];
void Add(int st,int ed,int num){
	tot++;
	G[tot]=node{ed,num,head[st]};
	head[st]=tot;
}
void DFS(int x,int fa){
	deep[x]=deep[fa]+1;
	pre[x]=fa;
	for(int i=head[x];i;i=G[i].last){
		int t=G[i].ed;
		if(t==fa)continue;
		bel[t]=G[i].num;
		DFS(t,x);
	}
}
void jump(int x,int y,int i){
	if(deep[x]<deep[y])swap(x,y);
	while(deep[x]>deep[y])P[i].push_back(bel[x]),x=pre[x];
	while(x!=y)P[i].push_back(bel[x]),P[i].push_back(bel[y]),x=pre[x],y=pre[y];
}
int main(){
	scanf("%d",&n);
	for(int x,y,i=1;i<=n-1;i++){
		scanf("%d %d",&x,&y);
		Add(x,y,i);
		Add(y,x,i);
	}
	DFS(1,0);
	scanf("%d",&m);
	for(int a,b,i=1;i<=m;i++){
		scanf("%d %d",&a,&b);
		jump(a,b,i);
	}
	for(int i=0;i<(1<<m);i++){
		memset(mark,false,sizeof(mark));
		int flag=0;
		for(int j=1;j<=m;j++){
			if((1<<(j-1))&i){
				flag++;
				const int sz=P[j].size();
				for(int k=0;k<sz;k++)mark[P[j][k]]=true;
			}
		}
		int sum=0;
		for(int j=1;j<=n-1;j++)sum+=(!mark[j]);
		if(flag&1)ans-=1LL<<sum;
		else ans+=1LL<<sum;
	}
	cout<<ans;
	return 0;
}