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: Wrong Answer)

import sys
import copy
sys.setrecursionlimit(1000000000)

def dfs(start, paths, total, usedIndices):
    if total <= 1:
        return 1
    
    indices = findNodeIndex(start, paths, usedIndices)
    if len(indices) == 0:
        return 0

    count = 0
    for i in indices:
        nextNode = paths[i][1] if paths[i][0] == start else  paths[i][0]
        count += dfs(nextNode, paths, total - 1, usedIndices + [i])
    return count

def findNodeIndex(node, paths, usedIndices):
    return [i for (i, row) in filter(lambda x: (not x[0] in usedIndices) and (x[1][0] == node or x[1][1] == node), [(i, row) for i, row in enumerate(paths)])]

N, M = map(int, input().split())
PATHS = [list(map(int, input().split())) for _ in range(M)]
#start = findNodeIndex(1, PATHS)

print(dfs(1, PATHS, N, []))