Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 51;
vector<int> adj[MAXN];
int d[MAXN], f[MAXN];
long long cur[MAXN], pre, x, y;

void dfs(int u, int dep, int p = 0){
	d[u] = dep;
	f[u] = p;
	for (int v : adj[u]){
		if (v != p){
			dfs(v, dep + 1, u);
		}
	}
}

void lca(int u, int v){
	if (d[u] < d[v]){
		swap(u, v);
	}
	while (d[u] > d[v]){
		x |= (1ll << u);
		u = f[u];
	}
	if (u == v){
		return;
	}
	while (u != v){
		x |= (1ll << u);
		y |= (1ll << v);
		u = f[u];
		v = f[v];
	}
}

int main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	long long  n;
	cin >> n;
	for (int i = 0; i < n - 1; ++i){
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1, 0);
	int m;
	cin >> m;
	long long ans = 0;
	for (int i = 0; i < m; ++i){
		int u, v;
		cin >> u >> v;
		x = 0;
		y = 0;
		lca(u, v);
		cur[i] = x ^ y;
	}
	int mx = (1 << m);
	for (int i = 0; i < mx; ++i){
		long long num = 0ll;
		int cnt = 0;
		for (int j = 0; j < m; ++j){
			if (i & (1 << j)){
				cnt++;
				num |= cur[j];
			}
		}
		ans += (1ll << (n - 1 - __builtin_popcountll(num))) * (cnt % 2 ? -1ll : 1ll);
	}
	cout << ans;
	return 0;
}