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 N 55
#define ll long long
#define pb push_back
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();
}
int n,m,fa[N],q[N],bk,state[N];
vector<int> v[N];
ll dp[N];
inline bool need_add(int i)
{
	int cnt=0;
	while(i)i-=i&-i,++cnt;
	return !(cnt&1);
}
int main()
{
	in(n);
	for(int i=1;i<n;++i)
	{
		int x,y;
		in(x);in(y);
		v[x].pb(y);
		v[y].pb(x);
	}
	q[bk++]=1;
	for(int i=0;i<n;++i)
	{
		int from=q[i];
		for(int j=0;j<v[from].size();++j)
		{
			int to=v[from][j];
			if(to==fa[from])continue;
			fa[to]=from;
			q[bk++]=to;
		}
	}
	in(m);
	for(int i=0;i<m;++i)
	{
		int x,y;
		in(x);in(y);
		state[x]^=1<<i;
		state[y]^=1<<i;
	}
	for(int i=n-1;i>=0;--i)
	{
		int from=q[i];
		for(int j=0;j<v[from].size();++j)
		{
			int to=v[from][j];
			if(to==fa[from])continue;
			state[from]^=state[to];
		}
	}
	ll ans=0;
	for(int mask=0;mask<(1<<m);++mask)
	{
		for(int i=n-1;i>=0;--i)
		{
			int from=q[i];
			dp[from]=1;
			for(int j=0;j<v[from].size();++j)
			{
				int to=v[from][j];
				if(to==fa[from])continue;
				if(mask&state[to])dp[from]*=dp[to];
				else dp[from]*=dp[to]<<1;
			}
		}
		if(need_add(mask))ans+=dp[1];
		else ans-=dp[1];
	}
	printf("%lld\n",ans);
	return 0;
}