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)

use proconio::{input, fastout};
use permutohedron::LexicalPermutation as _;

#[fastout]
fn main() {
    input!{
        n:usize,
        m:usize,
        ab: [(usize, usize); m]
    }
    let mut prm = (1..=n).collect::<Vec<usize>>();

    let mut path_num = 0;
    loop {
        let mut is_path = true;
        for i in 0..n-1 {
            let (v1, v2) = (prm[i], prm[i+1]);
            let mut edge_is_exist = false;
            for (a, b) in &ab {
                if (v1 == *a && v2 == *b) || (v1 == *b && v2 == *a) {
                    edge_is_exist = true;
                }
            }
            if !edge_is_exist {
                is_path = false;
                break;
            }
        }
        if is_path {
            path_num += 1;
        }
        if !prm.next_permutation() {
            break;
        }
        if prm[0] != 1 {
            break;
        }
    }
    println!("{}", path_num);
}