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 <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define all(v) v.begin(), v.end()
using namespace std;
using ll = long long;
using p = pair<int,int>;

int main() {
  int N ,M;
  cin >> N >> M;
  
  vector<vector<bool>> G(10, vector<bool>(10));
  rep (i, M)
  {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    G.at(a).at(b) = G.at(b).at(a) = true; 
  }
  
  vector<int> ord(N);
  rep (i, N)
  {
    ord.at(i) = i;
  }
  
  int count = 0;
  do {
    if (ord.at(0) != 0)
      break;
  	bool ok = true;
    for (int i = 0; i < N - 1; i++)
    {
      if(G.at(ord.at(i)).at(ord.at(i+1)) != true)
      {
        ok = false;
      }
    }
    if (ok == true)
    {
      count++;
    }
  } while (next_permutation(ord.begin(), ord.end()));
  cout << count << endl;
  
}