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<cstdio>
#include<algorithm>
#include<vector>
#define ll long long
using namespace std;
ll n, m;
ll xx[61];
ll dap;
ll gap;
pair<ll, ll> b[61];
pair<ll, ll> a[61];
ll parse[61][8];

vector < vector<pair<ll,ll>>> vt;
ll de[61];
ll dfs(ll here,ll par, ll en,ll ch2) {
	if (here == en)return 1;
	for (ll i = 0; i < vt[here].size(); i++) {
		ll next = vt[here][i].first;
		if (next == par)continue;
		if (dfs(next, here, en,ch2)) {
			xx[vt[here][i].second] |= ((ll)1<<ch2);
			return 1;
		}
	}
	return 0;
}
void func(ll here, ll ch,ll update) {
	if (here == m) {
		ll sum = 0;
		ll oo = 0;
		for (ll i = 0; i < n-1; i++) {
			if (ch & xx[i])oo++;
		}
		oo = (n - 1 - oo);
		sum = ((ll)1 << oo);
		if (update % 2)gap -= sum;
		else gap += sum;
		return;
	}
	func(here+1 , ch + ((ll)1 << here), update+1);
	func(here+1, ch, update);
}
int main()
{
	scanf("%lld", &n);
	vt.resize(n + 1);
	for (ll i = 0; i < n - 1; i++) {
		ll q, w;
		scanf("%lld %lld", &q, &w);
		b[i] = { q,w };
		vt[q].push_back({ w ,i});
		vt[w].push_back({ q ,i});
	}
	scanf("%lld", &m);
	for (ll i = 0; i < m; i++) {
		scanf("%lld %lld", &a[i].first, &a[i].second);
		dfs(a[i].first, -1, a[i].second, i);
	}
	func(0, 0, 0);
	printf("%lld", gap);
}