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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::Read;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let mut g = vec![vec![]; n];
    for k in 0..(n - 1) {
        let a: usize = it.next().unwrap().parse().unwrap();
        let b: usize = it.next().unwrap().parse().unwrap();
        g[a - 1].push((b - 1, k));
        g[b - 1].push((a - 1, k));
    }
    let mut q = vec![];
    let mut parent = vec![(0, 0); n];
    let mut depth = vec![0; n];
    let mut stack = vec![0];
    while let Some(v) = stack.pop() {
        q.push(v);
        if let Some(k) = g[v].iter().position(|u| u.0 == parent[v].0) {
            g[v].swap_remove(k);
        }
        for &(u, k) in g[v].iter() {
            parent[u] = (v, k);
            depth[u] = depth[v] + 1;
            stack.push(u);
        }
    }
    let m: usize = it.next().unwrap().parse().unwrap();
    let mut mask = vec![0];
    for _ in 0..m {
        let a: usize = it.next().unwrap().parse().unwrap();
        let b: usize = it.next().unwrap().parse().unwrap();
        let mut x = a - 1;
        let mut y = b - 1;
        let mut z = 0usize;
        while x != y {
            if depth[x] <= depth[y] {
                z |= 1 << parent[y].1;
                y = parent[y].0;
            } else {
                z |= 1 << parent[x].1;
                x = parent[x].0;
            }
        }
        let mut next = mask.clone();
        for mask in mask.into_iter() {
            next.push(mask | z);
        }
        mask = next;
    }
    let mut ans = 0i64;
    for (i, bit) in mask.iter().enumerate() {
        let c = i.count_ones();
        let sign = if c % 2 == 0 {1} else {-1};
        ans += sign * (1 << (n - 1 - bit.count_ones() as usize)) as i64;
    }
    println!("{}", ans);
}

fn main() {
    run();
}