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: Wrong Answer)

#include<stdio.h>

int pass[10][10] = {0} , count = 0 , n;
int flag[10] = {0} , a = 0;

int dfs(int start){

  flag[start] = 1;

  for(int i = 1;i <= n;i ++){
    if(flag[i] == 0){
      a = 1;
    }
  }

  if(a == 0){
    count ++;
  }
  a = 0;

  for(int i = 1;i <= n;i ++){
    if(pass[start][i] != 0 && flag[i] == 0){
      dfs(i);
    }
  }

  return count;
}

int main(void){
  int m;
  scanf("%d %d",&n ,&m);

  int a , b;
  for(int i = 0;i < m;i ++){
    scanf("%d %d",&a ,&b);
    pass[a][b] ++;
    pass[b][a] ++;
  }

  int ans = dfs(1);

  printf("%d\n",ans);

  return  0;
}