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 int long long
#define MN 55
int n,m,a[MN],b[MN],u[MN],v[MN],ok[MN],f[2][1<<21|1];
int fa[MN],dep[MN],ans;
vector<int>edge[MN];
void DFS(int x){
	for(int i=0;i<edge[x].size();++i)
		if(fa[x]!=edge[x][i]){
			fa[edge[x][i]]=x;
			dep[edge[x][i]]=dep[x]+1;
			DFS(edge[x][i]);
		}
}
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<n;++i){
		scanf("%lld%lld",&a[i],&b[i]);
		edge[a[i]].push_back(b[i]);
		edge[b[i]].push_back(a[i]);
	}
	scanf("%lld",&m);
	DFS(1);
	for(int i=1;i<=m;++i){
		scanf("%lld%lld",&u[i],&v[i]);
		if(dep[u[i]]<dep[v[i]])swap(u[i],v[i]);
		while(dep[u[i]]>dep[v[i]]) {
			ok[i]|=(1ll<<u[i]);
		//	cerr<<"shit "<<u[i]<<endl;
			u[i]=fa[u[i]];
		}
		while(u[i]!=v[i]){
			ok[i]|=(1ll<<u[i]),ok[i]|=(1ll<<v[i]);
		//	cerr<<"shit "<<u[i]<<" "<<v[i]<<endl;
			u[i]=fa[u[i]],v[i]=fa[v[i]];
		}
		ok[i]|=(1ll<<u[i]);
	}
	int ans=1;
	for(int i=1;i<n;++i)ans<<=1;
	f[0][0]=1;
	for(int i=1;i<n;++i){
		int qwq=0;
		for(int j=1;j<=m;++j)
			if((ok[j]&(1ll<<a[i]))&&(ok[j]&(1ll<<b[i]))) qwq|=(1ll<<j-1);
		for(int j=0;j<(1<<m);++j){
			f[i&1][j]+=f[!(i&1)][j];
			f[i&1][j|qwq]+=f[!(i&1)][j];
			f[!(i&1)][j]=0;
		}
	}
	printf("%lld\n",f[(n-1)&1][(1<<m)-1]);
	return 0;
}