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 (1.42.0) (judge verdict: Accepted)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use permutohedron::*;
use proconio::*;
use text_io::*;

const U_INF: usize = 1 << 60;
const I_INF: isize = 1 << 60;

fn main() {
    let n: usize = read!();
    let m: usize = read!();

    let mut v = vec![vec![]; n];

    for _ in 0..m {
        let xi: usize = read!();
        let yi: usize = read!();

        v[xi - 1].push(yi - 1);
        v[yi - 1].push(xi - 1);
    }

    let p = (0..n).permutations(n).collect_vec();

    let mut ans = 0;

    'par: for pi in p {
        if pi[0] != 0 {
            continue;
        }
        
        let mut now = pi[0];
        for &pii in &pi[1..] {
            if v[now].contains(&pii) {
                now = pii;
            } else {
                continue 'par;
            }
        }

        ans += 1;
    }

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