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: Accepted)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn vec<T, F: FnMut(usize) -> T>(n: usize, mut f: F) -> Vec<T> {
    (0..n).map(|i| f(i)).collect()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut neighbors: Vec<Vec<usize>> = vec![Vec::new(); n];
    for _ in 0..n - 1 {
        let a: usize = scan(tokens);
        let b: usize = scan(tokens);
        neighbors[a - 1].push(b - 1);
        neighbors[b - 1].push(a - 1);
    }
    let mut parent: Vec<usize> = (0..n).collect();
    let mut children: Vec<usize> = vec![0; n];
    fn dfs(
        i: usize,
        p: usize,
        neighbors: &Vec<Vec<usize>>,
        parent: &mut Vec<usize>,
        children: &mut Vec<usize>,
    ) {
        parent[i] = p;
        children[i] = 1 << i;
        for &j in &neighbors[i] {
            if j != p {
                dfs(j, i, neighbors, parent, children);
                children[i] |= children[j];
            }
        }
    }
    dfs(0, 0, &neighbors, &mut parent, &mut children);
    let m: usize = scan(tokens);
    let paths: Vec<usize> = vec(m, |_| {
        let u: usize = scan(tokens);
        let v: usize = scan(tokens);
        let mut u = u - 1;
        let mut v = v - 1;
        let mut p = 0;
        while children[u] & (1 << v) == 0 {
            p |= 1 << u;
            u = parent[u];
        }
        while u != v {
            p |= 1 << v;
            v = parent[v];
        }
        p
    });
    fn rec(paths: &[usize], set: usize, all: usize) -> i64 {
        if paths.is_empty() {
            1 << (all ^ set).count_ones()
        } else {
            rec(&paths[1..], set, all) - rec(&paths[1..], set | paths[0], all)
        }
    }
    println!("{}", rec(&paths, 0, (1 << n) - 2));
}