Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#define int long long
#define mod 1000000007
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
#define mp make_pair
using namespace std;
int get() { int r; cin >> r; return r; }
int n, m, key, p, ans;
int t[60][60], nm[30][60];
vector<int>v[60];
bool b[60];
bool f(int nd) {
	if (nd == key)return 1;
	b[nd] = 1;
	for (int i : v[nd]) {
		if (!b[i]) {
			if (f(i)) {
				nm[p][t[nd][i]] = 1;
				return 1;
			}
		}
	}
	b[nd] = 0;
	return 0;
}

signed main() {
	n = get();
	for1(i, n - 1) {
		int a, b; cin >> a >> b;
		t[a][b] = i; t[b][a] = i;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	m = get();
	for1(i, m) {
		int u; cin >> u >> key; p = i; f(u);
		for1(j, n) { b[j] = 0; }
	}
	int l = 1;
	for1(i, m)l *= 2;
	for0(i, l) {
		int ii = i, cnt = 0;
		for1(hoge, m) {
			if (ii % 2 == 1) {
				cnt++;
				for1(j, n - 1)if (nm[hoge][j])b[j] = 1;
			}
			ii /= 2;
		}
		p = 1;
		for1(j, n - 1) { if (b[j])b[j] = 0; else p *= 2; }
		if (cnt % 2)ans -= p;
		else ans += p;
	}
	cout << ans << endl;
}