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: Accepted)

//                             In The Name Of Allah
#include <bits/stdc++.h>
#define	ss second
#define ff first
#define use_fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define ret(n) return cout << n, 0
#define se(n) cout << setprecision(n) << fixed
#define pb push_back
#define int long long
#define ld long double
using namespace std; 
 
const int N = 60, OO = 1e9 + 100, M = 1e18 + 7, P = 6151, SQ = 120, lg = 21;
typedef pair <int, int> pii;
vector <pii> v[N];
bitset <N> mark[(1 << lg)];

int pw(int x, int y) {
	if(y == 0) return 1;
	int cnt = pw(x, y / 2);
	cnt = (cnt * cnt);
	if(y % 2 == 1)
		cnt = (cnt * x);
	return cnt;
}

bool dfs(int x, int en, int y, int msk) {
	if(en == x)
		return true;
	for(auto u : v[x]) {
		if(u.ff == y)	
			continue;
		bool b = dfs(u.ff, en, x, msk);
		if(b) {
			mark[msk][u.ss] = 1;
			return true;
		}
	}
	return false;
}

int32_t main() {
    use_fast;
    int n;
    cin >> n;
    for(int i = 0; i < n - 1; i++) {
    	int x, y;
    	cin >> x >> y;
    	v[x].pb({y, i});
    	v[y].pb({x, i});
	}
	vector <pii> q;
	int m;
	cin >> m;
	for(int i = 0; i < m; i++) {
		int x, y;
		cin >> x >> y;
		if(x == y)
			return cout << 0, 0;
		q.pb({x, y});
	}
	int ans = pw(2, n - 1);
	for(int i = 1; i < (1 << m); i++) {
		int cnt = 0;
		for(int j = 0; j < m; j++) {
			if((1 << j) & i) {
				cnt = j;
				break;
			}
		}
		mark[i] = mark[i - (1 << cnt)];
		dfs(q[cnt].ff, q[cnt].ss, 0, i);
		int t = __builtin_popcount(i);
		int h = n - 1 - mark[i].count();
		h = pw(2, h);
		if(t % 2 == 0) 
			ans = (ans + h);
		else
			ans = (ans - h);
	}
	cout << ans << endl;
	return 0; 
}
/*
be carefull :
1- if not solve after 20 min, read again twice
2- after submit read the code again
3- fun with contest
4- ... 
*/