Problem:
Score : 1000 points

Problem Statement

Inspired by the tv series Stranger Things, bear Limak is going for a walk between two mirror worlds.

There are two perfect binary trees of height H,
each with the standard numeration of vertices from 1 to 2^H-1.
The root is 1 and the children of x are 2 \cdot x and 2 \cdot x + 1.

Let L denote the number of leaves in a single tree, L = 2^{H-1}.

You are given a permutation P_1, P_2, \ldots, P_L of numbers 1 through L.
It describes L special edges that connect leaves of the two trees.
There is a special edge between vertex L+i-1 in the first tree and vertex L+P_i-1 in the second tree.

drawing for the first sample test, permutation P = (2, 3, 1, 4), special edges in green

Let's define product of a cycle as the product of numbers in its vertices.
Compute the sum of products of all simple cycles that have exactly two special edges, modulo (10^9+7).

A simple cycle is a cycle of length at least 3, without repeated vertices or edges.

Constraints

2 \leq H \leq 18

1 \leq P_i \leq L where L = 2^{H-1}

P_i \neq P_j (so this is a permutation)

Input

Input is given from Standard Input in the following format (where L = 2^{H-1}).

H
P_1 P_2 \cdots P_L

Output

Compute the sum of products of simple cycles that have exactly two special edges.
Print the answer modulo (10^9+7).

Sample Input 1

3
2 3 1 4

Sample Output 1

121788

The following drawings show two valid cycles (but there are more!) with products
2 \cdot 5 \cdot 6 \cdot 3 \cdot 1 \cdot 2 \cdot 5 \cdot 4 = 7200
and 1 \cdot 3 \cdot 7 \cdot 7 \cdot 3 \cdot 1 \cdot 2 \cdot 5 \cdot 4 \cdot 2 = 35280.
The third cycle is invalid because it doesn't have exactly two special edges.

Sample Input 2

2
1 2

Sample Output 2

36

The only simple cycle in the graph indeed has two special edges, and the product of vertices is
1 \cdot 3 \cdot 3 \cdot 1 \cdot 2 \cdot 2 = 36.

Sample Input 3

5
6 14 15 7 12 16 5 4 11 9 3 10 8 2 13 1

Sample Output 3

10199246

Solution in Rust (1.42.0) (judge verdict: Runtime Error)

use proconio::marker::*;
use proconio::*;

type Link = Option<Box<Node>>;

const F: usize = 26;

#[derive(Default)]
struct Node {
    cnt: [u32; F],
    next: [Link; F],
}

impl Node {
    fn insert(&mut self, s: &[u8]) -> (usize, usize) {
        if s.is_empty() {
            return (0, 0);
        }
        let k = (s[0] - b'a') as usize;
        let (mut bit, mut ans) = self.next[k]
            .get_or_insert_with(|| Box::new(Node::default()))
            .insert(&s[1..]);
        if s.len() == 1 {
            ans += self.cnt[k] as usize;
        }
        bit |= 1 << k;
        for (i, c) in self.cnt.iter_mut().enumerate() {
            *c += (bit >> i & 1) as u32;
        }
        (bit, ans)
    }
}

fn run() {
    input! {
        n: usize,
        mut s: [Bytes; n],
    }
    let mut root = Box::new(Node::default());
    s.sort_by_key(|s| s.len());
    let mut ans = 0;
    for s in s.iter_mut().rev() {
        s.reverse();
        ans += root.insert(s).1;
    }
    println!("{}", ans);
}

fn main() {
    let stack_size = 1 << 29;
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| run()).unwrap().join().unwrap();
}