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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
 
public class Main {
    static int n;
    static boolean[][] connected;
 
    static public int countPattern(int pos, Set<Integer> visited) {
        if (visited.size() == n) return 1;
 
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (connected[pos][i] && !visited.contains(i)) {
                visited.add(i);
                count += countPattern(i, visited);
                visited.remove(i);
            }
        }
        return count;
    }
 
    static public void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        int m = sc.nextInt();
 
        connected = new boolean[n][n];
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
 
            connected[a][b] = true;
            connected[b][a] = true;
        }
 
        Set<Integer> visited = new HashSet<>();
        visited.add(0);
        int count = countPattern(0, visited);
        System.out.println(count);
    }
}