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:
/*
Atcoder ABC152F
*/
#include <bits/stdc++.h>
using namespace std;

#define int long long

int n,m,g[55][55],t;
vector<int> e[55];
int found=0,u=0;
int f[55][1<<21];

void dfs(int x,int pre)
{
	if(x==t)
	{
		found = 1;
		return;
	}
	for(int y=1;y<=n;y++)
		if(g[x][y] && y!=pre)
		{
			dfs(y,x);
			if(found)
			{
				e[g[x][y]].push_back(u);
				return;
			}
		}
}

signed main()
{
	cin>>n;
	for(int i=1,x,y;i<n&&cin>>x>>y;i++)
		g[x][y] = g[y][x] = i;
	cin>>m;
	for(int i=1,x;i<=m&&cin>>x>>t;i++)
	{
		found = 0, u = i;
		dfs(x,0);
	}
	f[0][0] = 1;
	for(int i=0;i<n-1;i++)
		for(int s=0;s<=(1<<m)-1;s++)
		{
			f[i+1][s] += f[i][s];
			int t=s;
			for(int j=0;j<e[i+1].size();j++)
				t |= 1<<e[i+1][j]-1;
			f[i+1][t] += f[i][s];
		}
	cout<<f[n-1][(1<<m)-1]<<endl;
	
	return 0;
}