Problem:
Score : 300 points

Problem Statement

You are given an undirected unweighted graph with N vertices and M edges that contains neither self-loops nor double edges.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

How many different paths start from vertex 1 and visit all the vertices exactly once?

Here, the endpoints of a path are considered visited.

For example, let us assume that the following undirected graph shown in Figure 1 is given.

Figure 1: an example of an undirected graph

The following path shown in Figure 2 satisfies the condition.

Figure 2: an example of a path that satisfies the condition

However, the following path shown in Figure 3 does not satisfy the condition, because it does not visit all the vertices.

Figure 3: an example of a path that does not satisfy the condition

Neither the following path shown in Figure 4, because it does not start from vertex 1.

Figure 4: another example of a path that does not satisfy the condition

Constraints

2≦N≦8

0≦M≦N(N-1)/2

1≦a_i<b_i≦N

The given graph contains neither self-loops nor double edges.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the number of the different paths that start from vertex 1 and visit all the vertices exactly once.

Sample Input 1

3 3
1 2
1 3
2 3

Sample Output 1

2

The given graph is shown in the following figure:

The following two paths satisfy the condition:

Sample Input 2

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 2

1

This test case is the same as the one described in the problem statement.

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[doc = " Ported from [bluss/permutohedron](https://github.com/bluss/permutohedron)"]
pub trait LexicalPermutation {
#[doc = " Return \'true\' if the slice was permuted, \'false\' if it is already "]
#[doc = " at the last ordered permutation."]
    fn next_permutation(&mut self) -> bool;
#[doc = " Return \'true\' if the slice was permuted, \'false\' if it is already "]
#[doc = " at the first ordered permutation."]
    fn prev_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T]
where
    T: Ord,
{
#[doc = " Original author in Rust: Thomas Backman <serenity@exscape.org>"]
    fn next_permutation(&mut self) -> bool {
        if self.len() < 2 {
            return false;
        }
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] >= self[i] {
            i -= 1;
        }
        if i == 0 {
            return false;
        }
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i - 1] {
            j -= 1;
        }
        self.swap(j, i - 1);
        self[i..].reverse();
        true
    }
#[doc = " Original author in Rust: Thomas Backman <serenity@exscape.org>"]
    fn prev_permutation(&mut self) -> bool {
        if self.len() < 2 {
            return false;
        }
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] <= self[i] {
            i -= 1;
        }
        if i == 0 {
            return false;
        }
        self[i..].reverse();
        let mut j = self.len() - 1;
        while j >= i && self[j - 1] < self[i - 1] {
            j -= 1;
        }
        self.swap(i - 1, j);
        true
    }
}

fn get_line() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}

fn main() {
    let (n, m) = {
        let nums = get_line().split_whitespace()
            .map(|n| n.parse().unwrap())
            .collect::<Vec<usize>>();
        (nums[0], nums[1])
    };

    let mut edges: Vec<Vec<usize>> = Vec::new();
    for _ in 0..m {
        let ab = get_line().split_whitespace()
            .map(|n| n.parse().unwrap())
            .collect::<Vec<usize>>();
        edges.push(ab);
    }

    let mut order = (1..n+1).collect::<Vec<usize>>();
    let mut count = 0;

    loop {
        let mut ok = true;
        for i in 0..n-1 {
            let mut v = vec![order[i], order[i + 1]];
            v.sort();
            if !edges.contains(&v) { ok = false; }
        }
        if ok && order[0] == 1 { count += 1; }
        if !order.next_permutation() { break; }
    }

    println!("{}", count);
}