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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
/*
int count(int n, int m, int a[], int b[], int f){
    if (f == 1){
        return 0;
    }
    for (int i=0; i<m; i++){
        if (a[i] == n){
            count(b[i],m,a[i],b[i]);
        }else if (b[i] == n){
            count(a[i],m,a[i],b[i]);
        }
    }

}*/

int main(){
    int n,m;
    cin >> n >> m;
    int a[m],b[m];
    for (int i=0;i<m;i++){
        cin >> a[i] >> b[i];
    }
    int array[n-1];
    for (int i=0; i<n-1; i++){
        array[i] = i+2;
    }
    int root[n];
    root[0] = 1;
    int count = 0;
    do{
        for(int i=0; i<n-1; i++){
            root[i+1] = array[i];
        }
        for (int i=0; i<n-1; i++){
            int f = 0;
            for (int j=0; j<m; j++){
                /*cout << root[i];
                cout << a[j];
                cout << root[i+1];
                cout << b[j] << endl;*/
                if (root[i] == a[j]){
                    if (root[i+1] == b[j]){
                        f = 1;
                    }
                }else if (root[i] == b[j]){
                    if (root[i+1] == a[j]){
                        f = 1;
                    }
                }
                //cout << f << endl;
            }
            if (f == 0){
                break;
            }
            if (f == 1 && i == n-2){
                count++;
            }
        }
    }while(next_permutation(array,array+n-1));
    cout << count << endl;
    return 0;
}