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
#pragma GCC optimize ("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse4")
#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll
#define all(x) x.begin(), x.end()
#define trav(i,a) for(auto &i:a) 
inline int in(){int x;scanf("%lld",&x);return x;}
vector<pair<int,int>> g[51];
int val[25],mak=0;
#define f first
#define s second
bool dfs(int u,int par, int v)
{
	if(v==u)return 1;
	for(auto x:g[u])
	{
		if(x.f==par)continue;
		if(dfs(x.f,u,v))
		{
			mak|=(1LL<<x.s);
			return 1;
		}
	}
	return 0;
	
}
int32_t main()
{
	int n=in();
	int u,v;
	for(int i=0;i<n-1;i++)
	{
		u=in();v=in();
		g[u].push_back({v,i});
		g[v].push_back({u,i});
	}
	int m=in();
	for(int i=0;i<m;i++)
	{
		u=in();v=in();
		bool b=dfs(u,0,v);	
		val[i]=mak;
		mak=0;
	}
	
	int ans=0;
	
	for(int mask=0;mask<(1<<m);mask++)
	{
		//contains union of all edges in constraints in mask
		int submask=0;
		for(int j=0;j<m;j++)
		{
			if(mask&(1LL<<j))submask|=(val[j]);	
		}
		int num=__builtin_popcountll(submask);
		int rem=n-1-num;
		
		if(__builtin_popcountll(mask)&1)ans-=(1LL<<rem);
		else ans+=(1LL<<rem);
	}
	cout<<ans;
}