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<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int N,M;
vector<int>G[50];
int a[20],b[20];
int depth[50];
long subtree[50];
long dfs(int u,int p,int d)
{
	depth[u]=d;
	subtree[u]=1LL<<u;
	for(int v:G[u])
	{
		if(v!=p)
		{
			subtree[u]|=dfs(v,u,d+1);
		}
	}
	return subtree[u];
}
int u[50],v[50];
int check[50];
int main()
{
	cin>>N;
	for(int i=0;i<N-1;i++)
	{
		cin>>u[i]>>v[i];u[i]--,v[i]--;
		G[u[i]].push_back(v[i]);
		G[v[i]].push_back(u[i]);
	}
	dfs(0,-1,0);
	cin>>M;
	for(int i=0;i<M;i++)
	{
		cin>>a[i]>>b[i];
		a[i]--,b[i]--;
		long test=1LL<<a[i]|1LL<<b[i];
		for(int j=0;j<N-1;j++)
		{
			int U=depth[u[j]]<depth[v[j]]?v[j]:u[j];
			if((test&subtree[U])!=test&&(test&subtree[U])!=0)
			{
				check[j]|=1<<i;
			}
		}
	}
	long ans=0;
	for(int i=0;i<1<<M;i++)
	{
		int cnt=N-1;
		for(int j=0;j<N-1;j++)
		{
			if(check[j]&i)cnt--;
		}
		long now=1LL<<cnt;
		ans+=__builtin_parity(i)?-now:now;
	}
	cout<<ans<<endl;
}