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;
typedef long long int ll;
int n,m;
vector<int> adj[8];
unordered_set<string> ans;
void dfs(int i,vector<int> visited,int time){
    visited[i] = time;
    if(time == n){
        string str = "";
        for(int k=0;k<n;k++){
            str += to_string(visited[k]);
        }
        ans.insert(str);
        return;
    }
    for(auto j : adj[i]){
        if(visited[j] == -1){
            dfs(j,visited,time+1);
        }
    }

}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n>>m;
    for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;
        a--;b--;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    dfs(0,vector<int>(n,-1),1);
    cout<<ans.size()<<"\n";
    return 0;

}