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;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
ll n,m,t,ans,a[55],b[55],f[55],d[55],s[55];
vector<int> e[55];
void dfs(int x,int y){
	f[x]=y,d[x]=d[y]+1;
	for(int i=0;i<e[x].size();i++)
		if(e[x][i]!=y)dfs(e[x][i],x);
}
void gg(int x,int y){
	if(d[x]<d[y])swap(x,y);
	while(d[x]>d[y])
		s[x]=0,x=f[x];
	while(x!=y){
		s[x]=s[y]=0;
		x=f[x],y=f[y];
	}
}
int main(){
	n=read();
	for(int i=1;i<n;i++){
		int x=read(),y=read();
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0),m=read();
	for(int i=0;i<m;i++)
		a[i]=read(),b[i]=read();
	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))gg(a[j],b[j]);
		t=0;
		for(int j=2;j<=n;j++)
			t+=s[j];
		ans+=(1ll<<t)*(__builtin_popcount(i)&1?-1:1);
	}
	cout << ans;
	return 0;
}