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<cstdio>
#include<cstring>
typedef long long LL;
const int N=60;
int n,fa[N],fid[N],dep[N],m,head[N],cnt;
unsigned long long U[N],F[2][1<<20];
struct edge{
	int to,nxt,id;
}e[N<<1];
void dfs(int now){
	for(int i=head[now];i;i=e[i].nxt)
	if(!dep[e[i].to]){
		dep[e[i].to]=dep[now]+1;
		fa[e[i].to]=now,fid[e[i].to]=e[i].id;
		dfs(e[i].to);
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<n;++i){
		int u,v;
		scanf("%d%d",&u,&v);
		e[++cnt]=(edge){v,head[u],i},head[u]=cnt;
		e[++cnt]=(edge){u,head[v],i},head[v]=cnt;
	}
	dfs(dep[1]=1);
	scanf("%d",&m);
	for(int i=0;i<m;++i){
		int u,v;
		unsigned long long zt=0;
		scanf("%d%d",&u,&v);
		if(dep[u]<dep[v])u^=v^=u^=v;
		while(dep[u]!=dep[v])zt|=1uLL<<fid[u],u=fa[u];
		while(u!=v)zt|=(1uLL<<fid[u])|(1uLL<<fid[v]),u=fa[u],v=fa[v];
		U[i]=zt;
	}
	**F=1;
	for(int i=1;i<n;++i){
		memcpy(F[i&1],F[i&1^1],sizeof*F);
		for(int S=0;S<1<<m;++S){
			int zt=S;
			for(int j=0;j<m;++j)
			if(!(S>>j&1)&&(U[j]>>i&1))zt|=1<<j;
			F[i&1][zt]+=F[i&1^1][S];
		}
	}
	printf("%llu\n",F[n&1^1][(1<<m)-1]);
	return 0;
}