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;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<int,ll> pil;
typedef pair<ll,int> pli;
#define rep(i,n) for (int i=0;i<(n);++i)
#define REP(i,n) for (int i=1;i<=(n);++i)
#define all(x) x.begin(),x.end()
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define read(x) scanf("%d",&x)

int n,m;
vector<pii> G[55];
int add[50];
int dep[55],fa[55],faEdge[55];
inline void dfs(int u,int pa){
	dep[u]=dep[pa]+1;
	fa[u]=pa;
	for (int i=0;i<G[u].size();++i){
		int v=G[u][i].F;
		if (v!=pa) dfs(v,u);
		else faEdge[u]=G[u][i].S;
	}
}
ll dp[1<<20][50];
int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	rep(i,n-1){
		int x,y;cin>>x>>y;
		G[x].pb(mp(y,i));G[y].pb(mp(x,i));
	}
	dfs(1,0);
	cin>>m;
	rep(i,m){
		int x,y;cin>>x>>y;
		if (dep[x]<dep[y]) swap(x,y);
		while(dep[x]>dep[y]){
			add[faEdge[x]]|=1<<i;
			x=fa[x];
		}
		while(x!=y){
			add[faEdge[x]]|=1<<i;
			add[faEdge[y]]|=1<<i;
			x=fa[x];y=fa[y];
		}
	}
	dp[0][0]=1;
	rep(i,n-1) for (int mask=0;mask<(1<<m);++mask) if (dp[mask][i]){
		dp[mask][i+1]+=dp[mask][i];
		dp[mask|add[i]][i+1]+=dp[mask][i];
	}
	cout<<dp[(1<<m)-1][n-1]<<endl;
	
	return 0;
}