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 <stdio.h>
#include <vector>
#include <utility>
using namespace std;
typedef pair<int,int> pii;
vector<vector<pii> > v;
vector<pii> lst;
int N,M,S,E;
int visit[51];
vector<int> path[51][51];
int chk[21];
long long ans;
bool dfs(int x){
	if(x==E) return true;
	for(pii nx : v[x]){
		if(visit[nx.second]==0){
			visit[nx.second] = 1;
			if(dfs(nx.first)) return true;
			visit[nx.second] = 0;
		}
	}
	return false;
}
void bt(int x){
	if(x==M){
		int tmp[51]={0,}, edge=0,cnt=0;
		for(int i=0;i<M;i++){
			if(chk[i]){
				cnt++;
				for(int j : path[lst[i].first][lst[i].second]){
					tmp[j]=1;
				}
			}
		}
		for(int i=0;i<N-1;i++) edge+=tmp[i];
		ans += 1LL*(1-(cnt%2)*2)*(1LL<<(N-1-edge));
	}
	else{
		bt(x+1);
		chk[x]=1;
		bt(x+1);
		chk[x]=0;
	}
}
int main(){
	scanf("%d",&N);
	v.resize(N+1);
	for(int i=0;i<N-1;i++){
		int a,b;scanf("%d %d",&a,&b);
		v[a].push_back({b,i});
		v[b].push_back({a,i});
	}
	scanf("%d",&M);
	for(int i=0;i<M;i++){
		int a,b;
		scanf("%d %d",&a,&b);
		lst.push_back({a,b});
	}
	for(int i=1;i<=N;i++){
		for(int j=1;j<=N;j++){
			S=i;E=j;
			dfs(S);
			for(int k=0;k<N-1;k++){
				if(visit[k]) path[i][j].push_back(k);
				visit[k]=0;
			}
		}
	}
	bt(0);
	printf("%lld",ans);
}