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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int MOD = 1e9 + 7;
int n, m;
long long ans;
int v[25], u[25];
int a[60], b[60];
int dist[60][60];
vector<int> graph[60];

int k(int mask)
{
	if (__builtin_popcount(mask) % 2 == 0)
	    return 1;
	return -1;
}

void dfs(int v, int r, int p, int d)
{
	dist[r][v] = d;
	for (auto u : graph[v])
	    if (u != p) dfs(u, r, v, d + 1);
}

int on_path(int v, int a, int b)
{
	return dist[a][b] == dist[v][a] + dist[v][b];
}

int main()
{
	cin >> n;
	for (int i = 0; i < n - 1; i++)
	{
		int v, u;
		cin >> v >> u, v--, u--;
		a[i] = v, b[i] = u;
		graph[u].push_back(v);
		graph[v].push_back(u);
	}
	cin >> m;
	for (int i = 0; i < m; i++)
	{
		cin >> v[i] >> u[i], v[i]--, u[i]--;
	}
	ans = (1LL << n) / 2;
	for (int i = 0; i < n; i++)
	{
		dfs(i, i, n, 0);
	}
	for (int mask = 1; mask < (1 << m); mask++) {
		int c = 0;
		for (int i = 0; i < n - 1; i++)
		{
			int t = 0;
			for (int j = 0; j < m; j++)
			{
				if ((1 << j) & mask)
				if (on_path(a[i], v[j], u[j]))
				if (on_path(b[i], v[j], u[j]))
				t = 1;
			}
			c += t;
		}
		//cout << "!" << mask << " " << c << endl;
		ans = ans + k(mask) * (1LL << (n - 1 - c));
	}
	cout << ans;
}