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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#define INF 1000000009
#define LINF 1000000000000000009
#define all(a) a.begin(),a.end()
#define rep(i, n) for (ll i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;
template<class T> bool chmax(T &a, const T&b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T&b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll n, ll m) { return (m ? gcd(m, n%m) : n); }
ll lcm(ll n, ll m) { return n / gcd(n, m)*m; }

int pcnt(int n) {
	int ans = 0;
	for (int i = 0; i < 20; i++) {
		if ((1 << i)&n) ans++;
	}
	return ans;
}

vector<vector<int>> graph;
vector<vector<int>> st;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> a(n - 1), b(n - 1);
	graph.assign(n, vector<int>());
	vector<vector<int>> dis(n, vector<int>(n, INF));
	for (int i = 0; i < n; i++) dis[i][i] = 0;
	map<P, int> mp;
	for (int i = 0; i < n - 1; i++) {
		cin >> a[i] >> b[i];
		a[i]--; b[i]--;
		graph[a[i]].push_back(b[i]);
		graph[b[i]].push_back(a[i]);
		dis[a[i]][b[i]] = 1;
		dis[b[i]][a[i]] = 1;
		mp[{a[i], b[i]}] = i;
		mp[{b[i], a[i]}] = i;
	}
	for (int k = 0; k < n; k++) {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				chmin(dis[i][j], dis[i][k] + dis[k][j]);
			}
		}
	}
	/*for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (dis[i][j] == INF) cout << "INF" << ' ';
			else cout << dis[i][j] << ' ';
		}
		cout << endl;
	}*/
	int m;
	cin >> m;
	vector<int> u(m), v(m);
	st.assign(m, vector<int>());
	for (int i = 0; i < m; i++) {
		cin >> u[i] >> v[i];
		u[i]--; v[i]--;
		int np = u[i];
		while (np != v[i]) {
			for (int j : graph[np]) {
				if (dis[j][v[i]] < dis[np][v[i]]) {
					st[i].push_back(mp[{np, j}]);
					np = j;
					//cout << j << endl;
					continue;
				}
			}
		}
	}
	/*for (int i = 0; i < m; i++) {
		for (int j : st[i]) {
			cout << j << ' ';
		}
		cout << endl;
	}*/
	ll ans = 0;
	int sz = (1 << m);
	ans += (1LL << (n - 1));
	bitset<1100010> pcnts;
	for (int i = 0; i < sz; i++) {
		if (__builtin_popcount(i) & 1) pcnts[i] = 1;
	}
	for (int i = 1; i < sz; i++) {
		set<int> sts;
		for (int j = 0; j < m; j++) {
			if ((i >> j) & 1) {
				for (int k : st[j]) {
					if (!sts.count(k)) {
						sts.insert(k);
					}
				}
			}
		}
		int nm = sts.size();
		//cout << nm << endl;
		if (!(pcnts[i] & 1)) ans += (1LL << (n - 1 - nm));
		else ans -= (1LL << (n - 1 - nm));
		//cout << dp[i] << endl;
	}
	cout << abs(ans) << endl;
}