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)

N, M = map(int, input().split())

if M == 0:
    print(0)
    exit()

es = [[] for i in range(M)]
for i in range(M):
    a, b = map(int, input().split())
    a,b = a-1, b-1
    es[a].append(b)
    es[b].append(a)

def dfs(v, visited, es, cnt):

    #print("--------------")
    #print((v, visited, cnt))

    if cnt == N-1:
        return 1
    ret = 0
    for i in es[v]:
        if visited[i] == True:
            continue

        visited[i] = True
        ret += dfs(i, visited, es, cnt+1)
        visited[i] = False

    return ret

visited = [False] * N
visited[0] = True

ans = dfs(0,visited,es,0)
print(ans)