Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <LL,LL>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=1e9+7;

LL n,m,x,y,dp[60][1100000];
vector <pii> g[60];
vector <LL> contr[60],pass;

bool dfs(LL pos,LL no)
{
	if(pos==y) return true;
	rep(i,g[pos].size())
	{
		if(g[pos][i].fi==no) continue;
		pass.pb(g[pos][i].se);
		if(dfs(g[pos][i].fi,pos)) return true;
		pass.pop_back();
	}
	return false;
}

int main()
{
	cin>>n;
	rep(i,n-1)
	{
		cin>>x>>y;
		g[x].pb(mpr(y,i));
		g[y].pb(mpr(x,i));
	}
	cin>>m;
	rep(i,m)
	{
		cin>>x>>y;
		pass.clear();
		dfs(x,0);
		rep(j,pass.size()) contr[pass[j]].pb(i);
	}
	dp[0][0]=1;
	rep(i,n-1)
	{
		rep(j,1<<m)
		{
			if(dp[i][j]==0) continue;
			dp[i+1][j]+=dp[i][j];
			int tmp=j;
			rep(k,contr[i].size()) tmp|=(1<<contr[i][k]);
			dp[i+1][tmp]+=dp[i][j];
		}
	}
	cout<<dp[n-1][(1<<m)-1]<<endl;
	return 0;
}