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;
#define LL long long
struct Edge{
	int next,to;
}edge[100];
LL f[51][1<<20];
int fi[51],se=1,fa[51],fa1[51],d[51],dep[51];
inline void add_edge(int u,int v){edge[++se].next=fi[u],fi[u]=se,edge[se].to=v;}
inline void add(int u,int v){add_edge(u,v),add_edge(v,u);}
void dfs(int x){
	for(int i=fi[x];i;i=edge[i].next){
		int v=edge[i].to;if(v==fa[x])continue;
		fa[v]=x,fa1[v]=i/2,dep[v]=dep[x]+1,dfs(v);
	}
}
int main(){
	int n,u,v,m,k,s;scanf("%d",&n);
	for(int i=1;i<n;i++)scanf("%d%d",&u,&v),add(u,v);
	dfs(1);scanf("%d",&m);s=1<<m;
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		if(dep[u]>dep[v])swap(u,v);
		k=dep[v]-dep[u];
		for(int j=0;j<k;j++)d[fa1[v]]|=(1<<i),v=fa[v];
		while(v!=u)d[fa1[v]]|=1<<i,v=fa[v],d[fa1[u]]|=1<<i,u=fa[u];
	}
	f[1][0]=1;
	for(int i=1;i<n;i++){
		for(int j=0;j<s;j++){
			f[i+1][j]+=f[i][j];
			f[i+1][j|d[i]]+=f[i][j];
		}
	}
	printf("%lld",f[n][s-1]);
	return 0;
}