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: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define rep(i,n) for( int i = 0; i < n; i++ )
#define rrep(i,n) for( int i = n; i >= 0; i-- )
#define REP(i,s,t) for( int i = s; i <= t; i++ )
#define RREP(i,s,t) for( int i = s; i >= t; i-- )
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define INF 2000000000
#define mod 1000000007
#define INF2 1000000000000000000
#define int long long
vector<int> vec;
vector<int> G[51];
bool dfs(int u, int v, int p = -1) {
    //cout << u << " " << v << " " << p << endl;
    if(u == v) {
        vec.push_back(u);
        return true;
    }
    for(auto to: G[u]) {
        if (to == p) continue;
        if(dfs(to, v, u)) {
            vec.push_back(u);
            return true;
        }
    }
    return false;
}
map<pair<int, int>, int> mp;
signed main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; cin >> N;
    int a[51], b[51];
    rep(i, N - 1) {
        cin >> a[i] >> b[i];
        a[i]--; b[i]--;
        mp[{a[i], b[i]}] = i;
        mp[{b[i], a[i]}] = i;
        G[a[i]].push_back(b[i]);
        G[b[i]].push_back(a[i]);
    }
    int v[21], u[21];
    int M; cin >> M;
    rep(i, M) {
        cin >> v[i] >> u[i];
        v[i]--; u[i]--;
    }
    int ans = 0;
    rep(bit, (1 << M)) {
        int S = 0;
        rep(i, M) {
            if (bit & (1 << i)) {
                vec.clear();
                dfs(v[i], u[i]);
                rep(j, (int)(vec.size()) - 1) {
                    S |= (1LL << mp[{vec[j], vec[j + 1]}]);
                }
                //cout << "*" << vec.size() << endl;
            }
        }
        int cnt1 = __builtin_popcountll(bit);
        int cnt2 = __builtin_popcountll(S);
        if(cnt1 % 2 == 0) ans += (1LL << (N - 1 - cnt2));
        else ans -= (1LL << (N - 1 - cnt2));
        //cout << bitset<16>(bit) << " " << bitset<16>(S) << endl;
    }
    cout << ans << endl;


    return 0;
}