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;
typedef long long ll;

ll n;
vector<vector<ll> > adj(50, vector<ll>(50, -1ll));
vector<vector<ll> > paths(50, vector<ll>(50, 0ll));

ll pop_count(ll b) {
	ll cnt = 0;
	for (ll i = 0; i < 50; ++i) if ((b>>i)&1ll) cnt++;
	return cnt;
}

void dfs(ll x, ll p, ll bit) {
	for (ll v = 0; v < 50; ++v)
	{
		ll j = adj[x][v];
		if (j < 0) continue;
		if (v == p || paths[p][v] != 0) continue;
		paths[p][v] = bit + (1ll<<j);
		dfs(v, p, bit + (1ll<<j));
	}
	return;
}

int main() {
	cin >> n;
	for (ll i = 0; i < n-1; ++i)
	{
		ll a, b; cin >> a >> b;
		adj[a-1][b-1] = i;
		adj[b-1][a-1] = i;
	}
	for (ll i = 0; i < n; ++i) dfs(i, i, 0ll);

	ll m; cin >> m;
	vector<ll> cons(m);
	for (ll i = 0; i < m; ++i)
	{
		ll u, v;
		cin >> u >> v;
		cons[i] = paths[u-1][v-1]; 
	}
	vector<ll> uses(1ll<<m, 0);

	ll ans = (1ll<<(n-1ll));
	for (ll b = 1; b < (1ll<<m); ++b)
	{
		bool pop = pop_count(b)%2;
		ll cur = 0ll;
		while (((1ll<<cur) & b) == 0) cur++;
		ll use = uses[b - (1ll<<cur)] | cons[cur];
		uses[b] = use;
		if (pop) ans -= 1ll << (n - 1ll - pop_count(use));
		else ans += 1ll << (n - 1ll - pop_count(use));
	}

	cout << ans << endl;

	return 0;
}