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 <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int main() {
	int N;
	cin >> N;
	vector<vector<int> > G(N);
	for (int i = 0; i < N - 1; ++i) {
		int a, b;
		cin >> a >> b;
		--a, --b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int M;
	cin >> M;
	vector<int> p(M), q(M);
	for (int i = 0; i < M; ++i) {
		cin >> p[i] >> q[i];
		--p[i], --q[i];
	}
	vector<int> par(N), depth(N);
	function<void(int, int)> make_order = [&](int pos, int pre) {
		par[pos] = pre;
		for (int i : G[pos]) {
			if (i == pre) continue;
			depth[i] = depth[pos] + 1;
			make_order(i, pos);
		}
	};
	make_order(0, -1);
	vector<long long> A(1 << M);
	for (int i = 0; i < M; ++i) {
		long long x = 0;
		while (p[i] != q[i]) {
			if (depth[p[i]] < depth[q[i]]) {
				swap(p[i], q[i]);
			}
			x |= 1LL << (p[i] - 1);
			p[i] = par[p[i]];
		}
		for (int j = 1 << i; j < 2 << i; ++j) {
			A[j] = A[j - (1 << i)] | x;
		}
	}
	long long ans = 0;
	for (int i = 0; i < 1 << M; ++i) {
		long long mul = 1LL << (N - 1 - __builtin_popcountll(A[i]));
		if (__builtin_popcount(i) & 1) ans -= mul;
		else ans += mul;
	}
	cout << ans << endl;
	return 0;
}