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)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let N: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let (a, b): (Vec<usize>, Vec<usize>) = {
        let (mut a, mut b) = (vec![], vec![]);
        for _ in 0..(N - 1) {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            a.push(iter.next().unwrap().parse().unwrap());
            b.push(iter.next().unwrap().parse().unwrap());
        }
        (a, b)
    };
    let M: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let (u, v): (Vec<usize>, Vec<usize>) = {
        let (mut u, mut v) = (vec![], vec![]);
        for _ in 0..M {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            u.push(iter.next().unwrap().parse().unwrap());
            v.push(iter.next().unwrap().parse().unwrap());
        }
        (u, v)
    };

    let G = {
        let mut G = vec![std::collections::HashSet::new(); N + 1];
        for i in 0..(N - 1) {
            G[a[i]].insert(b[i]);
            G[b[i]].insert(a[i]);
        }
        G
    };

    let mut dp = vec![std::collections::HashSet::new(); M];
    for i in 0..M {
        let mut prev = vec![0; N + 1];
        let mut q = std::collections::VecDeque::new();
        q.push_back(u[i]);
        while let Some(j) = q.pop_front() {
            for &x in &G[j] {
                if prev[x] == 0 {
                    prev[x] = j;
                    q.push_back(x);
                }
            }
        }
        let mut j = v[i];
        while j != u[i] {
            dp[i].insert((std::cmp::min(j, prev[j]), (std::cmp::max(j, prev[j]))));
            j = prev[j];
        }
    }

    let mut res = 0i64;
    for i in 1..(1 << M) {
        let mut x = 0;
        let mut s = std::collections::HashSet::new();
        for j in 0..M {
            if (i >> j) & 1 == 1 {
                x += 1;
                for &y in &dp[j] {
                    s.insert(y);
                }
            }
        }
        let d = 1 << (N - 1 - s.len());

        if x % 2 == 1 {
            res += d;
        } else {
            res -= d;
        }
    }
    let ans = (1 << (N - 1)) as i64 - res;
    println!("{}", ans);
}