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 (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

#[fastout]
fn main() {
   input! {
      n:usize,
      ab:[(Usize1, Usize1); n-1],
      m:usize,
      uv:[(Usize1, Usize1); m]
   }

   let mut map = BTreeMap::new();
   map.insert((0, 0), 1);
   let mut to = vec![vec![]; n];
   for (i, &(a, b)) in ab.iter().enumerate() {
      to[a].push((b, i));
      to[b].push((a, i));
   }
   let mut eset = vec![0usize; m];
   for i in 0..m {
      let (u, v) = uv[i];
      let mut root = vec![];
      dfs(u, v, u, &to, &mut root);
      for j in root {
         eset[i] |= 1 << j;
      }
   }

   let mut ans: i64 = 0;
   for i in 0..1 << m {
      let mut mask = 0usize;
      let mut c: i64 = 1;
      for j in 0..m {
         if i >> j & 1 == 1 {
            mask |= eset[j];
            c *= -1;
         }
      }
      let white: usize = (0..n - 1).map(|i| mask >> i & 1).sum();
      ans += c * (1 << (n - 1 - white));
   }
   println!("{}", ans);
}

fn dfs(
   v: usize,
   tv: usize,
   p: usize,
   to: &Vec<Vec<(usize, usize)>>,
   root: &mut Vec<usize>,
) -> bool {
   if v == tv {
      true
   } else {
      for &(u, i) in to[v].iter() {
         if u == p {
            continue;
         }
         if dfs(u, tv, v, to, root) {
            root.push(i);
            return true;
         }
      }
      false
   }
}