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;
const int maxn = 50 + 10;
int n, m;
vector<int> E[maxn];
int dep[maxn], fa[maxn], x[maxn], y[maxn], cnt[maxn];
void dfs(int x)
{
	dep[x] = dep[fa[x]] + 1;
	for(auto v : E[x])
		if(v != fa[x])
		{
			fa[v] = x;
			dfs(v);
		}
}
void add(int u, int v, int val)
{
	if(dep[u] < dep[v]) 	
		swap(u, v);
	while(dep[u] != dep[v])
	{
		cnt[u] += val;
		u = fa[u];
	}
	while(u != v)
	{
		cnt[u] += val;
		cnt[v] += val;
		u = fa[u];
		v = fa[v];
	}
}	
ll solve(int now)
{
	if(now == m + 1)
	{
		int tot = 0;
		for(int i = 2; i <= n; ++i)
			if(!cnt[i])
				++tot;
		return 1ll << tot;
	}
	ll ret = solve(now + 1);
	add(x[now], y[now], 1);
	ret -= solve(now + 1);
	add(x[now], y[now], -1);
	return ret;
}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i < n; ++i)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		E[u].push_back(v);
		E[v].push_back(u);
	}
	dfs(1);
	scanf("%d", &m);
	for(int i = 1; i <= m; ++i)
		scanf("%d%d", &x[i], &y[i]);
	cout << solve(1) << endl;
	return 0;
}