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 Python (3.4.3) (judge verdict: Accepted)

n,m=map(int,input().split())
def dfs(now,depth):#現在見ている点、今まで列挙した頂点数
    if used[now]:
        return 0
    if depth==n:
        return 1
    used[now]=1#使用済みグラフを立てる
    ans=0#出力値を数える変数
    for i in range(n):#全部の遷移先を調べる
        if connect[now][i]:#nowから繋がっている頂点であれば遷移を試す。
            ans+=dfs(i,depth+1)
    used[now]=0#使用済みグラフを折る
    return ans
edge=[[int(i)for i in input().split()]for i in range(m)]#入力値受け取り、グラフの初期設定
connect=[[0 for i in range(n)]for i in range(n)]#接続状況を管理する配列
for i in range(m):
    connect[edge[i][0]-1][edge[i][1]-1]=1
    connect[edge[i][1]-1][edge[i][0]-1]=1
used=[0]*n#既にその頂点を訪問したか管理する配列
print(dfs(0,1))