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=[int(i) for i in input().split()]
pp=[[0 for i in range(2)] for j in range(m)]
not_bridge=0

#深さ優先探索---------------------------------
#参照元：Python3でグラフの実装＆深さ優先探索・幅優先探索の実装
#http://muromura.hatenablog.com/entry/2017/07/16/025414
def dfs(Graph, Start, Vn, Visited=None):
  
##追加（空から読んでも動くように修正）-----------------------
  if Graph[Start]==[] and Visited==None: 
        Next=Start+1
        dfs(Graph, Next, Vn, Visited)  #再帰
  else:
##--------------------------------------------------------

    if (Visited==None): 
        Visited = [] #開始点のときに探索済みリストをつくる
        Visited.append(Start)  #現在の頂点を探索済みリストに加える

    
    if len(Visited)==Vn: #全点訪問済みの場合、橋ではない
        #print(Visited)
        global not_bridge
        if len(set(Visited)) == n:
            not_bridge=not_bridge+1
#        print(Visited)
        return 0
        
    for Next in Graph[Start]:  #次に探索する頂点の候補
        Visited.append(Next)
        #if ( Next in Visited ):  continue  #訪問済みリストにある頂点は飛ばす
        dfs(Graph, Next, Vn, Visited)  #再帰
        Visited.pop()
    return 0
  
#--------------------------------------------------------        

for i in range(m) :
    pp[i] =[int(i) for i in input().split()]
    
#隣接リスト
l = [[] for i in range(n+1)] #頂点の数+1
for pi in pp:  #0-originedに注意、l[0]は空
    l[pi[0]].append(pi[1])
    l[pi[1]].append(pi[0])

#print(l)
dfs(l, 0, n)
print(not_bridge)