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;
typedef long long LL;
typedef pair<int, int> pii;
const int mod = 1e9+7;

vector<pii> g[55];
long long bits[1<<20];
int pow2[55];

bool dfs(int u, int p, int f, int j){
    for(int i=0;i<g[u].size();i++){
        int v = g[u][i].first, idx = g[u][i].second;
        if(v == p) continue;
        if(v == f){
            bits[1<<j] |= (1ll<<idx);
            return 1;
        }
        if(dfs(v, u, f, j)){
            bits[1<<j] |= (1ll<<idx);
            return 1;
        }
    }
    return 0;
}
int main(){
    int n, m;
    scanf("%d", &n);
    pow2[0] = 1;
    for(int i=1;i<=n;i++) pow2[i] = (pow2[i-1]*1ll*2)%mod;
    for(int i=0;i<n-1;i++){
        int a, b;
        scanf("%d%d", &a, &b);
        g[a].push_back(pii(b, i));
        g[b].push_back(pii(a, i));
    }
    scanf("%d", &m);
    for(int i=0;i<m;i++){
        int a, b;
        scanf("%d%d", &a, &b);
        dfs(a, 0, b, i);
    }
    for(int i=1;i<(1<<m);i++){
        if(__builtin_popcount(i) == 1) continue;
        for(int j=0;j<m;j++){
            if((i>>j)&1) bits[i] |= bits[1<<j];
        }
    }

    LL ans = pow2[n-1];
    for(int i=1;i<(1<<m);i++){
        int cnt = n-1-__builtin_popcountll(bits[i]);
        if(__builtin_popcount(i)%2 == 1){
            ans = (ans - pow2[cnt] + mod)%mod;
        }
        else{
            ans = (ans + pow2[cnt])%mod;
        }
    }
    printf("%lld\n", ans);
}