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;

vector < vector <int> > graph;
int mat[55][55];

int par[55];

void dfs(int u, int p = 0) {
	par[u] = p;
	for (int v : graph[u]) {
		if (v == p) {
			continue;
		}
		dfs(v, u);
	}
}

long long edgeMask[55];

void asign(int src, int des, int q) {
	dfs(src);
	int v = des;
	do {
		int u = par[v];
		int edge = mat[u][v];
		edgeMask[edge] |= (1LL << q);
		v = u;
	} while (v != src);
}

int n;
int m;

long long dp[51][(1LL << 20) + 5];

long long solve(int pos = 1, long long mask = 0) {
	if (pos >= n) {
		if (__builtin_popcountll(mask) == m) {
			return 1;
		}
		return 0;
	}
	long long &ret = dp[pos][mask];
	if (~ret) {
		return ret;
	}
	ret = solve(pos + 1, mask);
	ret += solve(pos + 1, mask | edgeMask[pos]);
	return ret;
}

signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);
	
	cin >> n;
	graph.resize(n + 1);
	for (int e = 1; e < n; e++) {
		int u, v;
		cin >> u >> v;
		graph[u].push_back(v);
		graph[v].push_back(u);
		mat[u][v] = mat[v][u] = e;
	}
	cin >> m;
	for (int q = 0; q < m; q++) {
		int u, v;
		cin >> u >> v;
		asign(u, v, q);
	}
	memset(dp, -1, sizeof dp);
	cout << solve() << endl;
}