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 N 55
#define pb push_back
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int>
#define ll long long
using namespace std;
inline void in(int &x)
{
	x=0;
	char c=getchar();
	while(c<'0'||c>'9')c=getchar();
	while(c>='0'&&c<='9')x=(x<<1)+(x<<3)+(c^48),c=getchar();
}
bool edge[N][N];
ll dp[N][1<<20];
int state[N];
int n,m,fa[N],dep[N];
void dfs(int from,int f)
{
	fa[from]=f;
	dep[from]=dep[f]+1;
	for(int to=1;to<=n;++to)
	if(to!=f&&edge[from][to])
	dfs(to,from);
}
void LCA(int x,int y,int i)
{
	while(x!=y)
	{
		if(dep[x]<dep[y])swap(x,y);
		state[x]|=1<<i;
		x=fa[x];
	}
}
int main()
{
	in(n);
	for(int i=1;i<n;++i)
	{
		int x,y;
		in(x);in(y);
		edge[x][y]=edge[y][x]=1;
	}
	dfs(1,0);
	in(m);
	for(int i=0;i<m;++i)
	{
		int x,y;
		in(x);in(y);
		LCA(x,y,i);
	}
	dp[2][0]=1;
	for(int i=2;i<=n;++i)
	for(int mask=0;mask<(1<<m);++mask)
	{
		dp[i+1][mask]+=dp[i][mask];
		dp[i+1][mask|state[i]]+=dp[i][mask];
	}
	printf("%lld\n",dp[n+1][(1<<m)-1]);
	return 0;
}