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 (3.8.2) (judge verdict: Wrong Answer)

#Nはノード数、Mはエッジ数
N, M = map(int, input().split())

path_matrix = []

#無向行列の初期化
for n in range(N):
    path_matrix.append([False] * N)

#まず無向グラフを表す行列を作る
for m in range(M):
    line = map(int, input().split())
    paths = [x - 1 for x in line]
    path_matrix[paths[0]][paths[1]] = True
    path_matrix[paths[1]][paths[0]] = True

#最初はvisited、つまり訪問したことのあるノードはひとつもないので、Falseにする
visited = [False] * N


#再帰関数、ここが難しいかもしれない。
def dfs(now, depth):
    if visited[now]:
        return 0
    if depth == N - 1:
        return 1

    visited[now] = True
    total_paths = 0
    for i in range(0, N):
        if path_matrix[now][i]:
            total_paths += dfs(i, depth + 1)


    return total_paths

print(dfs(0, 0))