Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

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

Sample Output 4

62

The tree in this input is shown below:

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
	int i, u, w, N, M, s[21], t[21], adj[51][51] = {};
	scanf("%d", &N);
	for (i = 1; i < N; i++) {
		scanf("%d %d", &u, &w);
		adj[u][w] = i;
		adj[w][u] = i;
	}
	scanf("%d", &M);
	for (i = 0; i < M; i++) scanf("%d %d", &(s[i]), &(t[i]));
	
	int P[21][51] = {}, prev[51], q[51], head, tail;
	for (i = 0; i < M; i++) {
		for (u = 1; u <= N; u++) prev[u] = -1;
		q[0] = s[i];
		prev[s[i]] = s[i];
		for (head = 0, tail = 1; head < tail; head++) {
			u = q[head];
			for (w = 1; w <= N; w++) {
				if (adj[u][w] == 0) continue;
				else if (w == t[i]) break;
				else if (prev[w] == -1) {
					prev[w] = u;
					q[tail++] = w;
				}
			}
			if (w <= N) break;
		}
		for (; w != u; w = u, u = prev[u]) P[i][adj[u][w]] = 1;
	}
	
	int j, sum[2], white[51];
	long long n, bit[51] = {1}, ans = (long long)1 << N - 1;
	for (i = 1; i <= N; i++) bit[i] = bit[i-1] << 1;
	for (n = 1; n < bit[M]; n++) {
		for (j = 1; j < N; j++) white[j] = 0;
		for (i = 0; i < M; i++) {
			if ((n | bit[i]) == n) {
				for (j = 1; j < N; j++) white[j] |= P[i][j];
			}
		}
		for (j = 1, sum[0] = 0; j < N; j++) sum[0] += white[j];
		for (i = 0, sum[1] = 0; i < M; i++) sum[1] += n / bit[i] % 2;
		if (sum[1] % 2 == 1) ans -= bit[N-1-sum[0]];
		else ans += bit[N-1-sum[0]];
	}

	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}