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>
using namespace std;
std::vector<pair<int,int>>edges[55] ;
vector<int>found,path;
void dfs(int a,int goal,int parent)
{
    if(a==goal)
    {
        found=path;
        return;
    }
    for(pair<int,int>e:edges[a])
    {
        if(e.first==parent)
        {
            continue;
        }
        path.push_back(e.second);
        dfs(e.first,goal,a);
        path.pop_back();
    }
}


int main() {
	// your code goes here
	int n;
	cin>>n;
	for(int i=0;i<n-1;i++)
	{
	    int a,b;
	    cin>>a>>b;
	    edges[a].emplace_back(b,i);
	    edges[b].emplace_back(a,i);
	}
	int m;
	cin>>m;
	vector<int>containme(n-1);
	for(int i=0;i<m;i++)
	{
	    int a,b;
	    cin>>a;
	    cin>>b;
	    found={-1};
	    dfs(a,b,-1);
	    if(found==vector<int>{-1})
	    {
	        cout<<"found";
	    }
	    
	    for(int e: found)
	    {
	        containme[e]|=(1<<i);
	    }
	    
	}
	vector<long long>dp(1<<m);
	dp[0]=1;
	for(int e=0;e<n-1;e++)
	{
	    vector<long long>new_dp(1<<m);
	    for(int mask=0;mask<(1<<m);mask++)
	    {
	        new_dp[mask]+=dp[mask];
	        new_dp[mask|containme[e]]+=dp[mask];
	    }
	    dp=new_dp;
	}
	cout<<dp[(1<<m)-1];
	
	
	return 0;
}
