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
#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define int long long
using namespace std;

int n,m;
vector<int> v[60],res[60];
int pw[60],val[2000010],ind[2000010];
int num[60][60];

inline bool dfs(int x,int pr,int des,int pos)
{
	res[pos].push_back(x);
	if(x==des){
		return true;
	}
	for(int i=0;i<v[x].size();i++) if(v[x][i]!=pr){
		int u=v[x][i];
		if(dfs(u,x,des,pos)) return true;
	}
	res[pos].pop_back();
	return false;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	pw[0]=1ll;
	for(int i=1;i<=n;i++) pw[i]=pw[i-1]*2ll;
	for(int i=1;i<n;i++){
		int x,y;
		cin>>x>>y;
		num[x][y]=i;
		num[y][x]=i;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	cin>>m;
	for(int i=1;i<=m;i++){
		ind[1ll<<(i-1ll)]=i;
	}
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		dfs(u,-1,v,i);
	}
	for(int i=1ll;i<(1ll<<m);i++){
		int mask=i&(-i);
		int loc=ind[mask];
		val[i]=val[i^mask];
		for(int j=0ll;j<res[loc].size()-1;j++){
			int pos1=res[loc][j],pos2=res[loc][j+1ll];
			val[i]|=(1ll<<(num[pos1][pos2]-1ll));
		}
	}
	int ans=0ll;
	for(int i=0ll;i<(1ll<<m);i++){
		int num=__builtin_popcountll(i);
		int tot=n-1ll-__builtin_popcountll(val[i]);
		if(num%2ll==0ll) ans+=pw[tot];
		else ans-=pw[tot];
	}
	cout<<ans<<endl;
	return 0;
}