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: Accepted)

def dfs(parent, connect, passed, N): #深さ優先探索
    cnt = 0
    passed.append(parent) #通った点をメモ
    if len(passed) == N: #全ての点を通過したかの判定
        cnt = 1
    for next_ in connect[parent]:
        if next_ in passed:
            continue
        cnt += dfs(next_, connect, passed, N) #全ての点を通るまで再帰. 戻り値の和で何通りかカウント
    passed.pop() #分岐点に戻る前にメモを1段だけ破棄
    return cnt    

N, M = map(int, input().split())
connect = [[] for _ in range(N)]
for _ in range(M):
    a, b = map(int, input().split())
    connect[a-1].append(b-1)
    connect[b-1].append(a-1)
ans = dfs(0, connect, [], N) #点0からのdfs
print(ans)