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;
#define pb push_back
#define pf push_front
typedef long long lint;
typedef complex<double> P;
#define mp make_pair
#define fi first
#define se second
typedef pair<int,int> pint;
#define All(s) s.begin(),s.end()
#define rAll(s) s.rbegin(),s.rend()
#define REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)
//問題文および制約はちゃんと確認しよう！
//サイズは10^5じゃなくて2×10^5とかかもしれないし、重要な制約?条件を見落としているかも
//とりあえずサンプルを読んでから解法を考えよう？
int mask[58];
vector<pint> gr[58];
lint dp[52][(1<<20)+10];
lint dfs(int v,int u,int g){
	if(v==g) return 0;
	rep(i,gr[v].size()){
		pint p=gr[v][i];
		if(p.fi==u) continue;
		lint x=dfs(p.fi,v,g);
		if(x>=0L) return x+(1LL<<p.se);
	}
	return -1;
}
int main()
{
	int n,a,b,m;
	cin>>n;
	rep(i,n-1){
		cin>>a>>b;gr[a].pb(mp(b,i));gr[b].pb(mp(a,i));
	}
	cin>>m;
	memset(mask,false,sizeof(mask));
	rep(i,m){
		cin>>a>>b;
		lint ma=dfs(a,-1,b);
		rep(j,n-1){
			if((ma&(1LL<<j))) mask[j]+=(1<<i);
		}
		//cout<<ma<<endl;
	}
	//rep(i,n-1) cout<<i<<' '<<mask[i]<<endl;
	memset(dp,0,sizeof(dp));
	dp[0][0]=1;
	rep(i,n-1) rep(j,(1<<m)){
		dp[i+1][j]+=dp[i][j];
		dp[i+1][(j|mask[i])]+=dp[i][j];
	}
	cout<<dp[n-1][(1<<m)-1]<<endl;
}

