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;

#ifdef DGC
#include "debug.h"
#else
#define debug(...) 9715
#endif
typedef long long ll;
typedef long double ld;
typedef complex<double> point;
#define F first
#define S second



int main()
{
	#ifdef DGC
		//freopen("a.in", "r", stdin);
		//freopen("b.out", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	int n;
	cin >> n;
	vector<vector<pair<int, int>>> adj(n);
	for (int i = 1, u, v; i < n; ++i)
	{
		cin >> u >> v;
		--u, --v;
		adj[u].push_back({ v, i });
		adj[v].push_back({ u, i });
	}

	vector<vector<int>> p(n, vector<int>(n));
	vector<vector<int>> e(n, vector<int>(n));
	for (int i = 0; i < n; ++i)
	{
		function<void(int, int)> dfs = [&](int u, int a)
		{
			for (auto v : adj[u])
				if (v.F != a)
				{
					dfs(v.F, u);
					e[i][v.F] = v.S;
					p[i][v.F] = u;
				}
		};
		p[i][i] = e[i][i] = -1;
		dfs(i, -1);
	}

	int m;
	cin >> m;
	vector<int> r(n);
	for (int i = 0; i < m; ++i)
	{
		int u, v;
		cin >> u >> v;
		--u, --v;
		while (v != u)
		{
			r[e[u][v]] |= 1<<i;
			v = p[u][v];
		}
	}

	vector<ll> dp(1<<m);
	dp[0] = 1;
	for (int i = 1; i < n; ++i)
		for (int j = 1<<m; --j >= 0; )
			dp[j|r[i]] += dp[j];
	cout << dp[(1<<m)-1] << "\n";

	return 0;
}
