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
import java.util.*;
public class Main{
	static ArrayList<Edge>[] map;
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		map=new ArrayList[n];
		for(int i=0; i<n; i++){
			map[i]=new ArrayList<>();
		}
		for(int i=1; i<n; i++){
			int a=sc.nextInt()-1;
			int b=sc.nextInt()-1;
			map[a].add(new Edge(b,i));
			map[b].add(new Edge(a,i));
		}
		
		int m=sc.nextInt();
		long[] s=new long[m];
		for(int i=0; i<m; i++){
			int u=sc.nextInt()-1;
			int v=sc.nextInt()-1;
			s[i]=dfs(u,v,-1);
		}
		long ans=0;
		for(int i=0; i<1<<m; i++){
			long tmp=0;
			for(int j=0; j<m; j++){
				if((i>>j & 1)==1){
					tmp|=s[j];
				}
			}
			long pm=Long.bitCount(tmp);
			if(Integer.bitCount(i)%2==0){
				ans+=1L<<(n-1-pm);
			}else{
				ans-=1L<<(n-1-pm);
			}
		}
		System.out.println(ans);
	}
	static long dfs(int now,int goal,int front){
		for(Edge i:map[now]){
			if(i.to==front){
				continue;
			}
			long tmp=dfs(i.to,goal,now);
			if(tmp!=-1){
				return tmp+(1L<<i.id);
			}
		}
		if(now==goal){
			return 0;
		}else{
			return -1;
		}
	}
	static class Edge{
		int to,id;
		Edge(int to,int id){
			this.to=to;
			this.id=id;
		}
	}
}
