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
#include <bits/stdc++.h>

using namespace std;
#define int long long
const int MOD = 1000000007;

vector<vector<pair<int, int> > > g;
vector<int> B;
void dfs(int a, int par, int d) {
	B[a] = d;
	for (auto p : g[a]) {
		if (p.first != par) {
			dfs(p.first, a, d ^ ((int)1 << p.second));
		}
	}
}
int pcnt(int a) {
	int res = 0;
	while (a > 0) {
		res += a % 2;
		a /= 2;
	}
	return res;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	g.resize(N);
	B.resize(N);
	int res = 0;
	for (int i = 0; i < N - 1; i++) {
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		g[a].emplace_back(b, i);
		g[b].emplace_back(a, i);
	}
	dfs(0, -1, 0);
	int M;
	cin >> M;
	vector<int> C(M);
	for (int i = 0; i < M; i++) {
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		C[i] = (B[a] ^ B[b]);
	}
	for (int i = 0; i < (1 << M); i++) {
		int x = 0;
		int cnt = 1;
		for (int j = 0; j < M; j++) {
			if (i & (1 << j)) {
				cnt = -cnt;
				x |= C[j];
			}
		}
		//cerr << i << " " << pcnt(x) << endl;
		res += cnt * ((int)1 << ((N - 1) - (pcnt(x))));
	}
	cout << res << endl;
}