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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

	static ArrayList<Integer>[] adj;
	static int timer, par[], tin[], tout[];

	static int lca(int u, int v) {
		while (true) {
			if (tin[v] >= tin[u] && tin[v] <= tout[u])
				break;
			u = par[u];
		}
		return u;
	}

	static int[] prefix;

	static void update(int u, int lca) {
		prefix[u]++;
		prefix[lca]--;
	}

	static int dfs(int u, int p) {
		par[u] = p;
		tin[u] = timer++;
		int ans = 0;
		for (int v : adj[u])
			if (v != p) {
				ans += dfs(v, u);
				prefix[u] += prefix[v];
			} 
		if (prefix[u] > 0)
			ans++;
		tout[u] = timer - 1;
		return ans;
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		adj = new ArrayList[n];
		par = new int[n];
		tin = new int[n];
		prefix = new int[n];

		tout = new int[n];
		for (int i = 0; i < n; i++) {
			adj[i] = new ArrayList();
		}
		for (int i = 1; i < n; i++) {
			int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
			adj[u].add(v);
			adj[v].add(u);
		}
		dfs(0, 0);
		
		int[][] lca = new int[n][n];
		int m = sc.nextInt();
		int[] a = new int[m], b = new int[m];
		for (int i = 0; i < m; i++) {
			a[i] = sc.nextInt() - 1;
			b[i] = sc.nextInt() - 1;
			lca[a[i]][b[i]] = lca(a[i], b[i]);
		}
		long ans = 0;
		for (int msk = 0; msk < 1 << m; msk++) {
			Arrays.fill(prefix, 0);
			for (int i = 0; i < m; i++) {
				if ((msk & 1 << i) == 0)
					continue;
				update(a[i], lca[a[i]][b[i]]);
				update(b[i], lca[a[i]][b[i]]);

			}
			int c = dfs(0, 0);  

			ans += (1L << (n - 1 - c)) * (Integer.bitCount(msk) % 2 == 0 ? 1 : -1);
		}
		out.println(ans);
		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

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

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}

}