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;
#define LL long long
#define pb push_back
#define chmin(a,b) a=min(a,b)
const int N=55;

int n,m;
int a[N],b[N];
int dist[N][N];
int task[N];
vector<int> e[N];
LL dp[1<<20];

int main()
{
	
	cin>>n;
	memset(dist,60,sizeof(dist));
	for(int i=1;i<n;i++)
	{
		int u,v;
		cin>>u>>v;
		e[u].pb(v); e[v].pb(u);
		dist[u][v]=dist[v][u]=1;
		a[i]=u; b[i]=v;
	}
	for(int i=1;i<=n;i++)
		dist[i][i]=0;
	for(int k=1;k<=n;k++)
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
		chmin(dist[i][j],dist[i][k]+dist[k][j]);
	cin>>m;
	for(int i=0;i<m;i++)
	{
		int u,v; cin>>u>>v;
		for(int j=1;j<n;j++)
			if(dist[u][a[j]]+dist[b[j]][v]==dist[u][v]-1 ||\
			dist[v][a[j]]+dist[b[j]][u]==dist[u][v]-1) task[j]|=1<<i;
	}
	dp[0]=1;
	for(int i=1;i<n;i++)
	for(int mask=(1<<m)-1;mask>=0;mask--)
		dp[mask|task[i]]+=dp[mask];
	cout<<dp[(1<<m)-1]<<'\n';
	
	return 0;
}