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>
#define ll long long
using namespace std;
const int N=55;
vector<int> G[N];
bitset<N> s[N][N];
int pre[N],S,E;
int a[N],b[N];
map<pair<int,int>,int> ma;
void dfs(int u,int fa) {
	if(u==E) {
		int t=E;
		while(pre[t]!=-1) s[S][E]|=(1LL<<ma[{t,pre[t]}]),t=pre[t];
		return;
	}
	for(int i=0;i<G[u].size();i++) {
		int v=G[u][i];
		if(v==fa) continue;
		pre[v]=u;
		dfs(v,u);
	}
}
int main() {
	int n,m;
	scanf("%d",&n);
	for(int i=1,u,v;i<n;i++) {
		scanf("%d%d",&u,&v);
		ma[{v,u}]=ma[{u,v}]=i;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=n;j++) {
			if(i==j) continue;
			S=i,E=j,pre[i]=-1;
			dfs(i,-1);
		}
	}
	scanf("%d",&m);
	for(int i=0;i<m;i++) {
		scanf("%d%d",&a[i],&b[i]);
	}
	ll ans=1LL<<(n-1);
	for(int i=1;i<1<<m;i++) {
		int num=0;
		bitset<N> t(0);
		for(int j=0;j<m;j++) {
			if((1<<j)&i) num++,t|=s[a[j]][b[j]];
		}
		if(num&1) ans-=1LL<<(n-1-t.count());
		else ans+=1LL<<(n-1-t.count());
	}
	printf("%lld\n",ans);
	return 0;
}