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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

#define rep for(int i = 0;i<n;i++)
const int MOD = 1000000007;

int n,m, ans;
bool graph[n][n],frag[n];

void dfs(bool a,int b,int c){
        if(b == n){
                ans += 1;  return;
        }
        for(int i = 0;i<n;i++){
                if(a[i] == false and graph[c][i] == true){
                    a[i] = true;  dfs(a,b+1, i);
                }
        }
}

int main(){
     cin >> n >> m;
     for(int i = 0;i<m;i++){
             int ai,bi;cin >> ai >> bi;
             graph[ai-1][bi-1] = true;graph[bi-1][ai-1] = true;
     }
     ans = 0;
     frag[0] = true;
     dfs(frag, 1, 0);

     cout << ans << endl;
}