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.*;

class Main {
	Scanner sc;
	int N, M;
	List<int[]>[] to;
	long[] p;
	
	@SuppressWarnings("unchecked")
	private void calc() {
		sc = new Scanner(System.in);
		N = sc.nextInt();
		to = new List[N];
		for (int i = 0; i < N; i++) to[i] = new ArrayList<>();
		for (int i = 0; i < N-1; i++) {
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			to[a].add(new int[] {b, i});
			to[b].add(new int[] {a, i});
		}
		M = sc.nextInt();
		p = new long[M];
		for (int i = 0; i < M; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			p[i] = pathToLong(u, v);
		}

		long s = 0;
		for (int i = 0; i < (1<<M); i++) {
			long c = 0;
			for (int j = 0; j < M; j++) {
				if ( (i&(1<<j)) > 0 ) c |= p[j];
			}
			int bitCount = Integer.bitCount(i);
			s += ((bitCount%2 == 0)?1:-1)*(1L << (N-1-Long.bitCount(c)));
		}
		
		System.out.println(s);
	}

	private long pathToLong(int u, int v) {
		return dfs(u, u, v);
	}

	private long dfs(int src, int curr, int dest) {
		for (int[] v : to[curr]) {
			if (v[0] == src) continue;
			if (v[0] == dest) return (1L << v[1]);
			long f = dfs(curr, v[0], dest);
			if (f > 0) return f | (1L << v[1]);
		}
		return 0;
	}
	
	public static void main(String[] args) {
		new Main().calc();
	}
}
