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>
using namespace std;
#define _GLIBCXX_DEBUG //これつけるとA[N]でもいいらしい
//for文のマクロ
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define big 1000000007
#define all(a) sort((a).begin(),(a).end()) //ソートのマクロ
#define Re(a) reverse((a).begin(),(a).end())
#define YN(a) if(a){cout<<"Yes"<<endl;}else cout<<"No"<<endl;//条件によってYes、Noを出力する
bool graph[8][8]={};
int n,m;
int cnt=0;
void dfs(int now,vector<int> visited){
    visited[now]=true;
    bool all_visited=true;
    rep(i,n){
        if(!visited[i]){
            all_visited = false;
        }
    }
    if(all_visited){
        cnt++;//すべて訪問済みの場合のみ、条件を満たしている
        return ;
    }
    rep(i,n){
        if(!visited[i] && graph[now][i]){
            //頂点nowとiが結ばれており、iが未訪問ならばさらに探索する
            //cout<<now<<"->"<<i<<endl;
            dfs(i,visited);
        }
    }
}
int main(){
    cin>>n>>m;
    vector<int> visited(n,false);
    rep(i,m){
        int a,b;
        cin>>a>>b;
        a--,b--;
        graph[a][b]=graph[b][a]=true;
    }
    dfs(0,visited);
    cout<<cnt<<endl;
}