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 INF 1000000000
#define LINF 1000000000000000000
#define MOD 1000000007
#define mod 998244353
#define INF63 1061109567
#define INF127 9187201950435737471
#define UINF 18446744073709551615
#define F first
#define S second
#define ll long long
#define N 55
#define M 1200010
using namespace std;
ll n,m,cons[N],mask,dp[N][M];
bool fd;
vector<pair<ll,ll> > vt[N];
void dfs(ll x,ll des,ll lst)
{
	ll i;
	if(x==des)
	{
		fd=true;
		return;
	}
	for(i=0;i<vt[x].size();i++)
	{
		if(vt[x][i].F!=lst)
		{
			dfs(vt[x][i].F,des,x);
			if(fd)
			{
				mask|=((ll)1<<vt[x][i].S);
				return;
			}
		}
	}
	return;
}
int main(){
	ll i,j,x,y;
	cin>>n;
	for(i=0;i+1<n;i++)
	{
		cin>>x>>y;
		x--;
		y--;
		vt[x].push_back(make_pair(y,i));
		vt[y].push_back(make_pair(x,i));
	}
	memset(cons,0,sizeof(cons));
	cin>>m;
	for(i=0;i<m;i++)
	{
		cin>>x>>y;
		x--;
		y--;
		fd=false;
		mask=0;
		dfs(x,y,-1);
		for(j=0;j+1<n;j++)
		{
			if(mask&((ll)1<<j))
			{
				cons[j]|=((ll)1<<i);
			}
		}
	}
	memset(dp,0,sizeof(dp));
	dp[0][0]=1;
	for(i=0;i+1<n;i++)
	{
		for(j=0;j<(1<<m);j++)
		{
			dp[i+1][j]+=dp[i][j];//white
			dp[i+1][(j|cons[i])]+=dp[i][j];//black
		}
	}
	cout<<dp[n-1][(1<<m)-1]<<endl;
	return 0;
}