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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <algorithm>
#include <vector>
#include <numeric>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define all(a) (a).begin(), (a).end()

int N, M;
bool adj[10][10];
int ans;

int main()
{
  scanf( "%d%d", &N, &M );
  rep( i, M )
  {
    int a, b;
    scanf( "%d%d", &a, &b );
    --a; --b;
    adj[a][b] = adj[b][a] = true;
  }

  std::vector<int> perm( N-1 );
  std::iota( all(perm), 1 );

  do
  {
    bool fl = true;

    rep( i, N-1 )
    {
      int prv = i ? perm[i-1] : 0;

      fl &= adj[prv][perm[i]];
    }

    if( fl )
      ++ans;
  } while( std::next_permutation(all(perm)) );

  printf( "%d\n", ans );
  
  return 0;
}