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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

public class Main {
	static List<List<Hen>> list;

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		list = new ArrayList<>(n);
		for (int i = 0; i < n; i++) {
			list.add(new ArrayList<>());
		}
		for (int i = 0; i < n - 1; i++) {
			String[] sa = br.readLine().split(" ");
			int a = Integer.parseInt(sa[0]) - 1;
			int b = Integer.parseInt(sa[1]) - 1;
			list.get(a).add(new Hen(b, i));
			list.get(b).add(new Hen(a, i));
		}

		int m = Integer.parseInt(br.readLine());
		long[] cond = new long[m];
		for (int i = 0; i < m; i++) {
			String[] sa = br.readLine().split(" ");
			int u = Integer.parseInt(sa[0]) - 1;
			int v = Integer.parseInt(sa[1]) - 1;
			List<Integer> route = new ArrayList<>();
			dfs(u, v, -1, route);
			for (int j : route) {
				cond[i] += 1L << j;
			}
		}
		br.close();

		long ans = 0;
		int m2 = 1 << m;
		for (int i = 0; i < m2; i++) {
			long conds = 0;
			for (int j = 0; j < m; j++) {
				if ((i >> j & 1) == 1) {
					conds |= cond[j];
				}
			}
			int white = Long.bitCount(conds);
			long val = 1L << (n - 1 - white);
			int num = Integer.bitCount(i);
			if (num % 2 == 0) {
				ans += val;
			} else {
				ans -= val;
			}
		}
		System.out.println(ans);
	}

	static boolean dfs(int x, int to, int p, List<Integer> route) {
		if (x == to) {
			return true;
		}
		for (Hen h : list.get(x)) {
			if (h.v == p) {
				continue;
			}
			if (dfs(h.v, to, x, route)) {
				route.add(h.i);
				return true;
			}
		}
		return false;
	}

	static class Hen {
		int v, i;

		public Hen(int v, int i) {
			this.v = v;
			this.i = i;
		}
	}
}