Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

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

Sample Output 4

62

The tree in this input is shown below:

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using int64 = long long;
using uint64 = unsigned long long;

using namespace std;

struct Edge {
    int id;
    int to;
};

constexpr int MAX_N = 50;

vector<Edge> tree[MAX_N];

int64 enumerateAffectedEdges(int u, int par, int goal) {
    int64 res = 0;
    for (const Edge& e: tree[u]) {
        if (e.to == par) continue;
        if (e.to == goal) {
            res |= (1LL << e.id);
        } else {
            int64 subres = enumerateAffectedEdges(e.to, u, goal);
            res |= subres;
            if (subres > 0) res |= (1LL << e.id);
        }
    }
    return res;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    int n;
    cin >> n;

    for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        tree[a].push_back((Edge){ i, b });
        tree[b].push_back((Edge){ i, a });
    }

    int m;
    cin >> m;

    vector<int64> affected_edges(m);
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        affected_edges[i] = enumerateAffectedEdges(u, -1, v);
    }

    int64 ans = 0;
    for (int mask = 0; mask < (1 << m); mask++) {
        int cnt = 0;
        int64 union_affected_edges = 0;
        for (int i = 0; i < m; i++) {
            if ((mask >> i) & 1) {
                cnt++;
                union_affected_edges |= affected_edges[i];
            }
        }
        int white = __builtin_popcount(union_affected_edges);
        if (cnt % 2 == 1) {
            ans -= (1LL << (n - 1 - white));
        } else {
            ans += (1LL << (n - 1 - white));
        }
    }
    cout << ans << endl;

    return 0;
}