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: Runtime Error)

q=[]
qq=[]
def dfs(j,q,icnt,count,n,qq):
    if len(qq)==n:
        count+=1
#        print("1",n,j,icnt,count,q[j],qq)
        return count
    icnt+=1
    
    for j2 in q[j]:
        if not j2 in qq:
            qq.append(j2)
#            print("2",n,j,icnt,count,q[j],qq)
            count=dfs(j2,q,icnt,count,n,qq)
            qq.pop()
#        print("3",n,j,icnt,count,q[j],qq)
    return count

n,m=map(int,input().split())
q=[[0]*2 for i in range(m)]
for i in range(m):
    q[i]=list(map(int,input().split()))

#n=3
#m=3
#ab=[[1,2],[1,3],[2,3]]

#n=7
#m=7
#ab=[[1,3],[2,7],[3,4],[4,5],[4,6],[5,6],[6,7]]

q=[[]*2 for i in range(n)]
for i in range(m):
    if not q[ab[i][1]-1] in q[ab[i][0]-1]:
        q[ab[i][0]-1].append(ab[i][1]-1)    
    if not q[ab[i][0]-1] in q[ab[i][1]-1]:
        q[ab[i][1]-1].append(ab[i][0]-1)    
#print(q)    

count=0
j=0
qq=[j]
icnt=1
count=dfs(j,q,icnt,count,n,qq)
print(count)