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 namespace std;
#define MOD 1000000007
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define show(x) for(auto i: x){cout << i << " ";}
typedef long long ll;

struct Edge{
    int to, id;
    Edge(int to, int id): to(to), id(id){} 
};

vector<Edge> edge[52];
//vector<int> a    :int要素を格納するまっさらな動的配列
//vector<int> a(5) :予め5個のint要素を確保した動的配列
//vector<int> a[5] :まっさらな動的配列を5つ確保している静的配列

bool fds(vector<int>& root, int now, int targ, int parent){
    if (targ == now){
        return true;
    }
    for (auto next : edge[now]){
        if (next.to == parent) continue;
        if (fds(root, next.to, targ, now)){
            root.push_back(next.id);
            return true;
        }
    }
    return false;
}

int main(){
    int n, m;
    cin >> n;
    
    rep(i, n-1){
        int tmpa, tmpb; cin >> tmpa >> tmpb;
        tmpa--; tmpb--;
        edge[tmpa].emplace_back(tmpb, i);
    }    

    cin >> m;
    vector<ll> bitmap_range(m);
    rep(i, m){
        int tmpa, tmpb; cin >> tmpa >> tmpb;
        tmpa--; tmpb--;
        vector<int> root;
        fds(root, tmpa, tmpb, -1);
        for(auto hen:root){
            bitmap_range[i] += 1ll << hen;            
        }
    }
    int ans = 0;
    rep(i, 1<<m){
        ll score;
        ll sum_range = 0;
        rep(j, m){
            if (i>>j&1) sum_range |= bitmap_range[j];
        }
        score = 1<<(n-1-__builtin_popcountll(sum_range));
        if (__builtin_popcountll(i) == 0){
            ans += score;
        } else {
            ans -= score;
        }
    }

    cout << ans << endl;
}