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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#define NMAX 8

int  a[NMAX+1][NMAX+1] = { 0 };
int  n, c = 0;

void dfs(int);

int main(void)
{
  int  m, i, j, k; 

  scanf("%d %d", &n, &m);
  for( i = 1; i <= m; i++){
    scanf("%d %d", &j, &k);
    a[j][k] = 1;
    a[k][j] = 1;
  }

  a[1][0] = 1;
  dfs(1);

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

  return 0;
}

void dfs(int v)
{
  int  i, t=0;

  for(i=1; i<=n; i++){
    t += a[i][0];
  }
  if(t == n){
    c++;
    return;
  }

  for( i = 1; i <= n; i++){
    if(a[i][0] == 0 && a[v][i] == 1){
      a[i][0] = 1;
      dfs( i );
      a[i][0] = 0;
    }
  }
  return;
}