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>
typedef long long ll;
using namespace std;
const ll mod = 1000000007;
const ll mod998 = 998244353;
const ll intmax = 2147483647;
const ll llmax = 9223372036854775807;
const char sp = ' ';

struct edge {
	int to, id;
};

int N, M, ednum[50];
vector<edge>v[50];
ll DP[1 << 20];

bool dfs(int x, int goal, int from, int t) {
	if (x == goal)return true;
	for (edge &e : v[x]) {
		if (e.to == from)continue;
		if (dfs(e.to, goal, x, t)) {
			ednum[e.id] += t;
			return true;
		}
	}
	return false;
}

void dfs2(int x, int from) {
	for (edge &e : v[x]) {
		if (e.to == from)continue;
		dfs2(e.to, x);
		for (int i = (1 << M) - 1; i >= 0; i--) {
			DP[i | ednum[e.id]] += DP[i];
		}
	}
}

int main() {
	cin >> N;
	for (int i = 0; i < N - 1; i++) {
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		v[a].push_back({ b,i });
		v[b].push_back({ a,i });
	}
	cin >> M;
	for (int i = 0; i < M; i++) {
		int u, v;
		cin >> u >> v;
		dfs(u - 1, v - 1, -1, 1 << i);
	}
	DP[0] = 1;
	dfs2(0, -1);
	cout << DP[(1 << M) - 1] << endl;
}