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)

# 頂点および辺の数を入力
N, M = map(int, input().split())
# 辺のリストを作成
list_edge = [[] for _ in range(N + 1)]
for _ in range(M):
    a, b = map(int, input().split())
    list_edge[a].append(b)
    list_edge[b].append(a)
# パス（リスト）のリストを初期化
list_path = [[1]]
# パスが存在する、かつパスの長さが頂点の数未満である場合に繰り返し
while list_path != [] and len(list_path[0]) < N:
    # パスのリストから先頭のパスを取り出す
    path = list_path.pop(0)
    # パスの最後尾の頂点から訪問可能な頂点について、、
    for node in list_edge[path[-1]]:
        # その頂点にまだ訪問していない場合、、
        if node not in path:
            # パスのリストに新たなパスを加える
            list_path.append(path + [node])
# パスの数を出力する
print(len(list_path))