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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdlib>
#include <map>
#include <iomanip>
#include <set>
#include <stdio.h>
#include <ctype.h>
#include <random>
#include <string.h>
using namespace std;
#define pie 3.141592653589793238462643383279
#define mod 1000000007
#define all(vec) vec.begin(),vec.end()

int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x%y);
}
int lcm(int x, int y) {
	return x / gcd(x, y)*y;
}
int kai(int x, int y) {
	int res = 1;
	for (int i = x - y + 1; i <= x; i++) {
		res *= i; res %= mod;
	}
	return res;
}
int mod_pow(int x, int y, int m) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % m;
		}
		x = x * x % m;
		y >>= 1;
	}
	return res;
}

int comb(int x, int y) {
	if (y > x)return 0;
	return kai(x, y) * mod_pow(kai(y, y), mod - 2, mod) % mod;
}

long long n, m;
vector<pair<int, int>> vec[55];
int d[55];
long long p[55];
int g[55];
long long cnt[25];
long long ans;

void dfs(int v, int pp,int k) {
	for (int i = 0; i < (int)vec[v].size(); i++) {
		if (vec[v][i].first == pp)continue;
		p[vec[v][i].first] = vec[v][i].second;
		d[vec[v][i].first] = k + 1;
		g[vec[v][i].first] = v;
		dfs(vec[v][i].first, v, k + 1);
	}
}

signed main() {
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int a, b; cin >> a >> b;
		vec[a].push_back(make_pair(b, i));
		vec[b].push_back(make_pair(a, i));
	}
	dfs(1, 0, 0);
	cin >> m;
	for (int i = 0; i < m; i++) {
		int a, b; cin >> a >> b;
		while (d[a] < d[b]) {
			cnt[i] += 1 << p[b];
			b = g[b];
		}
		while (d[a] > d[b]) {
			cnt[i] += 1 << p[a];
			a = g[a];
		}
		while (a != b) {
			cnt[i] += 1 << p[b];
			cnt[i] += 1 << p[a];
			b = g[b];
			a = g[a];
		}
	}
	for (int i = 0; i < (1 << m); i++) {
		long long cnt2 = 0, cnt3 = 0;
		for (long long j = 0; j < m; j++)if (i&(1 << j)) {
			cnt2 |= cnt[j];
			cnt3++;
		}
		long long cnt4 = 0;
		for (long long j = 0; j < n - 1; j++)if (cnt2&(1 << j))cnt4++;
		if (cnt3 % 2)ans -= (1 << (n - 1 - cnt4));
		else ans += (1 << (n - 1 - cnt4));
	}
	cout << ans << endl;
}