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;

int n, m, x[55];
vector<pair<int, int>> v[55];
long long dp[1 << 20];

bool dfs1(int a, int b, int c, int d) {
	if (a == c)
		return 1;
	for (auto& i : v[a])
		if (i.first != b && dfs1(i.first, a, c, d)) {
			x[i.second] ^= 1 << d;
			return 1;
		}
	return 0;
}

void dfs2(int a, int b) {
	
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		int ta, tb;
		scanf("%d%d", &ta, &tb);
		v[ta].emplace_back(tb, i);
		v[tb].emplace_back(ta, i);
	}
	scanf("%d", &m);
	for (int i = 0; i < m; i++) {
		int ta, tb;
		scanf("%d%d", &ta, &tb);
		dfs1(ta, 0, tb, i);
	}
	dp[0] = 1;
	// for (int i = 1; i < n; i++) printf("x[%d]=%d\n", i, x[i]);
	for (int i = 1; i < n; i++)
		for (int j = (1 << m) - 1; j >= 0; j--)
			dp[j | x[i]] += dp[j];
	printf("%lld\n", dp[(1 << m) - 1]);
}