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;
int n,m,xh;
bool vv[55];
long long ans=0;
vector<pair<int,int> >v[55];
vector<int>pth,tpth[22];
map<pair<int,int>,int>mp;
void find_path(int x,int y,int needtofind)
{
	if(tpth[xh].size()!=0)
	{
		return;
	}
	if(x==needtofind)
	{
		tpth[xh]=pth;
		return;
	}
	for(int i=0;i<v[x].size();i++)
	{
		int z=v[x][i].first;
		if(z==y)
		{
			continue;
		}
		pth.push_back(v[x][i].second);
		find_path(z,x,needtofind);
		pth.pop_back();
	}
}
signed main()
{
	cin>>n;
	for(int i=0;i<n-1;i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);
		a--;
		b--;
		v[a].push_back(make_pair(b,i));
		v[b].push_back(make_pair(a,i));
	}
	cin>>m;
	for(int i=0;i<m;i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);
		a--;
		b--;
		pth.clear();
		find_path(a,-1,b);
		xh++;
	}
	for(int i=1;i<(1<<m);i++)
	{
		memset(vv,false,sizeof(vv));
		int x=__builtin_popcount(i),yyy=n-1;
		for(int j=0;j<m;j++)
		{
			if((i>>j)&1)
			{
				for(int k=0;k<(int)tpth[j].size();k++)
				{
					if(!vv[tpth[j][k]])
					{
						vv[tpth[j][k]]=true;
						yyy--;
					}
				}
			}
		}
		if(x%2==1)
		{
			ans-=(1ll<<yyy);
		}
		else
		{
			ans+=(1ll<<yyy);
		}
	}
	cout<<ans+(1ll<<(n-1))<<endl;
	return 0;
}