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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
input = sys.stdin.readline

def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))
import itertools

def main():
    mod=10**9+7
    N,M=MI()
    adj=[[] for _ in range(N)]
    for _ in range(M):
        a,b=MI()
        a-=1
        b-=1
        adj[a].append(b)
        adj[b].append(a)
        
        
    ans=0
    for ite in itertools.permutations(range(1,N), N-1):
        v=0
        f=1
        for nv in ite:
            if nv in adj[v]:
                v=nv
            else:
                f=0
                break
        ans+=f
        
    print(ans)
        
            
    

main()