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.*;
import java.io.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };

	static int n, m, con[][];
	static List<Integer> l[];
	static long memo[];
	static Map<String,Integer> mp; 
	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		n = fs.nextInt();
		l = new ArrayList[n];
		mp = new HashMap<>();
		for(int i=0;i<n;i++)l[i] = new ArrayList<>();
		for(int i=0;i<n-1;i++) {
			int a = fs.nextInt()-1, b = fs.nextInt()-1;
			l[a].add(b);
			l[b].add(a);
			mp.put(Integer.toString(a)+"to"+Integer.toString(b), i);
			mp.put(Integer.toString(b)+"to"+Integer.toString(a), i);			
		}
		m = fs.nextInt();
		con = new int[m][2];
		for(int i=0;i<m;i++) {
			int u = fs.nextInt()-1, v = fs.nextInt()-1;
			con[i][0] = u;
			con[i][1] = v;
		}
		memo = new long[m];//各区間の通る点の情報をbitでもつ（頂点３２個以上なのでlong）
		
		for(int i=0;i<m;i++) {
			dfs(i,con[i][0],con[i][1],-1);
		}
		
		long res = 0; //どこかしらの区間が全部白になる場合の数
		//全白にするパスをbit全探索する
		for(int i=1;i<(1<<m);i++) {
			int cnt = 0;
			long bit = 0;
			for(int j=0;j<m;j++) {
				if((i>>j&1)==1) {
					cnt++;
					bit |= memo[j];
				}
			}
			int num = Long.bitCount(bit);
			if(cnt%2==0) res -= Math.pow(2, n-1-num);
			else res += Math.pow(2, n-1-num);
			
		}
		long ans = (long) (Math.pow(2, n-1) - res);
		System.out.println(ans);
	}
	static boolean dfs(int i, int u, int v, int parent) {
		if(u == v) {
			return true;
		}
		for(int to:l[u]) {
			if(to==parent)continue;
			if(dfs(i, to, v, u)) {
				int edge = mp.get(Integer.toString(u)+"to"+Integer.toString(to)); 
				memo[i] |= 1L<<edge;
				return true;
			}
		}
		return false;
	}
}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}