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: Accepted)


#pragma GCC optimize ("O3")

#include <bits/stdc++.h>
#define FOR(i, m, n) for (int i = (m); i < (n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define dump(x)  cerr << #x << " = " << (x) << endl
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl

using namespace std;

typedef long long ll;
typedef pair<ll,ll> Pll;
typedef pair<int,int> Pin;

ll INF = 10e16;
int inf = 10e8;

template<typename T>
ostream& operator << (ostream& os, const vector<T>& vec) {
    os << "[ ";
    REP(i, vec.size()) {
        os << vec[i] << " ";
    }
    os << "]";

    return os;
}

int _id(int a, int b, int N) {
    if (a < b) return N * a + b;
    return N * b + a;
}

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    
    int N;
    cin >> N;

    int E[N][N];
    REP(i, N) REP(j, N) {
        if (i == j) E[i][j] = 0;
        else E[i][j] = inf;
    }

    vector<int> P[N][N];
    REP(i, N - 1) {
        int a, b;
        cin >> a >> b;
        --a; --b;
        E[a][b] = 1;
        E[b][a] = 1;
        P[a][b].eb(_id(a, b, N));
        P[b][a].eb(_id(b, a, N));
    }

    REP(k, N) {
        REP(i, N){
            REP(j, N){
                if (E[i][j] > E[i][k] + E[k][j]){
                    E[i][j] = E[i][k] + E[k][j];

                    P[i][j].reserve(P[i][k].size() + P[k][j].size());
                    P[i][j].insert(P[i][j].end(), P[i][k].begin(), P[i][k].end());
                    P[i][j].insert(P[i][j].end(), P[k][j].begin(), P[k][j].end());
                }
            }
        }
    }

    int M;
    cin >> M;

    ll res = 1 << (N-1);

    int u[M], v[M];
    REP(i, M) {
        cin >> u[i] >> v[i];
        --u[i]; --v[i];
    }

    ll ans = 0;
    REP(i, 1 << M) {
        int p[N * N];
        REP(i, N * N) p[i] = 0;

        int s = 0;
        for(int j=i, k=0; j > 0; j >>= 1, ++k) {
            if(j & 1) {
                for (auto e: P[u[k]][v[k]]) {
                    p[e] |= 1;
                }
                ++s;
            }
        }

        int cnt = 0;
        REP(j, N * N) cnt += p[j];

        if (s % 2) ans -= pow(2, N - cnt -1);
        else ans += pow(2, N - cnt -1);
    }


    cout << fixed << setprecision(20) << ans << endl;

}