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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#ifdef DEBUG
#define debug(...) __VA_ARGS__
#else
#define endl '\n'
#define debug(...)
#endif
#define debugf(...) debug(printf(__VA_ARGS__))

const int N = 54;
using ull = unsigned long long;
using ll = long long;

using ii = pair<int, int>;
vector<ii> g[N];

int P[N], Pid[N];
int dep[N];
void build(int u, int p=0, int pid=0) {
    dep[u] = dep[p] + 1;
    P[u] = p;
    Pid[u] = pid;
    for (auto [v, id] : g[u]) if (v != p) {
        build(v, u, id);
    }
}

ull ct[N];
void add(int u, int v, int i) {
    auto upd = [&](int x) {
        ct[i] |= 1ull << Pid[x];
    };
    while (dep[u] > dep[v]) {
        upd(u);
        u = P[u];
    }
    while (dep[v] > dep[u]) {
        upd(v);
        v = P[v];
    }
    while (u != v) {
        upd(u); upd(v);
        u = P[u]; v = P[v];
    }
}

int (((((main)))))() <%
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    int n; cin >> n;
    for (int i = 0; i < n-1; i++) {
        int u, v;
        cin >> u >> v;
        g[u].emplace_back(v, i);
        g[v].emplace_back(u, i);
    }

    build(1);

    int m; cin >> m;
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        add(u, v, i);
    }

    ll ans = 0;
    const ll mod = 1e9 + 7;
    for (int msk = 0; msk < (1<<m); msk++) {
        ll include = __builtin_popcount(msk) & 1 ? -1 : 1;
        ull edges = (1ll<<(n-1)) - 1;
        for (int b = 0; b < m; b++) {
            if (msk & (1<<b))
                edges &= ~ct[b];
        }
        debugf("msk=%x, edge_count=%d\n", msk, __builtin_popcount(edges));
        // ll f = n - __builtin_popcount(edges) - 1;
        ll f = __builtin_popcount(edges);
        ans += include * (1ll << f) % mod;
        ans = (ans%mod + mod) % mod;
    }

    cout << ans << endl;

    return 0;
%>