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 vi = std::vector<int>;
using vpi = std::vector<std::pair<int, int>>;
using vvpi = std::vector<vpi>;

int N;
vvpi edges;
vi conditions;
std::vector<bool> visited;

bool find(int, int, int);

int main()
{
	scanf("%d", &N);
	edges.resize(N);
	conditions.resize(N - 1);
	for (int i{}; i < N - 1; i++)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		a--; b--;
		edges[a].push_back({b, i});
		edges[b].push_back({a, i});
	}
	int M;
	scanf("%d", &M);
	visited.resize(N);
	for (int i{}; i < M; i++)
	{
		std::fill(visited.begin(), visited.end(), false);
		int u, v;
		scanf("%d%d", &u, &v);
		u--; v--;
		find(u, v, i);
	}
	// for (auto& e: conditions) printf("%d ", e);
	// putchar('\n');
	int dp_size{1 << M};
	std::vector<int64_t> dp(dp_size);
	dp[0] = 1;
	for (auto& e: conditions)
		for (int i{dp_size - 1}; i >= 0; i--)
			dp[i | e] += dp[i];
	printf("%lld\n", dp.back());

	return 0;
}

bool find(int index, int key, int condition_i)
{
	if (visited[index]) return false;
	visited[index] = true;
	if (index == key) return true;
	for (auto& e: edges[index])
	{
		if (!find(e.first, key, condition_i)) continue;
		conditions[e.second] |= (1 << condition_i);
		return true;
	}
	return false;
}
