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>
#include <cstring>
using namespace std;
typedef pair<int, int> ii;
int N, M, masks[49];
ii uv[20];
long long dptable[1 << 20][49];
vector<ii> adjlist[50];
bool dfs(int v, int p, int mask) {
	if (v == uv[mask].second - 1) return true;
	for (ii i: adjlist[v]) {
		if (i.first == p) continue;
		if (dfs(i.first, v, mask)) {
			masks[i.second] = masks[i.second] | (1 << mask);
			return true;
		}
	}
	return false;
}
long long dp(int i, int j) {
	if (j == N - 1) return i == (1 << M) - 1 ? 1: 0;
	if (dptable[i][j] != -1) return dptable[i][j];
	return dptable[i][j] = (dp(i | masks[j], j + 1) + dp(i, j + 1));
}
int main() {
	memset(dptable, -1, sizeof dptable);
	int a, b;
	scanf("%d", &N);
	for (int i = 1; i < N; ++i) {
		scanf("%d %d", &a, &b);
		adjlist[a - 1].push_back(ii(b - 1, i - 1));
		adjlist[b - 1].push_back(ii(a - 1, i - 1));
	}
	scanf("%d", &M);
	for (int i = 0; i < M; ++i) scanf("%d %d", &uv[i].first, &uv[i].second);
	for (int i = 0; i < M; ++i) dfs(uv[i].first - 1, -1, i);
	printf("%lld", dp(0, 0));
	return 0;
}