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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(){
		int n,m,i,ab[30][2],ans=0,ttyk[10][10]={0},j;
		scanf("%d%d",&n,&m);
		for(i=0;i<m;i++){
				scanf("%d%d",&ab[i][0],&ab[i][1]);
				ttyk[ab[i][0]][ab[i][1]]=1;
				ttyk[ab[i][1]][ab[i][0]]=1;
		}
		int dfs[100000][10]={0},st=0,fn=1;
		dfs[0][0]=1;
		dfs[0][1]=1;
		while(fn>st){
				int f=0;
				for(j=1;j<=n;j++){
						if(ttyk[dfs[st][0]][j]==1&&dfs[st][j]!=1){
								dfs[fn][0]=j;
								for(i=1;i<=n;i++)dfs[fn][i]=dfs[st][i];
								dfs[fn][j]=1;
								fn++;
								f=1;
						}
				}
				if(f==0){
						for(j=1;j<=m;j++){
								if(dfs[st][j]==0)break;
						}
						if(j==m+1)ans++;
				}
				st++;
		}
		printf("%d\n",ans);
		return 0;
}