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)

import numpy as np
import math

# K = int(input())
# A, B = map(int, input().split())
# A = list(map(int, input().split()))

N, M = map(int, input().split())

edges = []
E = np.zeros((N, N), dtype=int)

for m in range(M):
    s, t = map(int, input().split())
    s = s - 1
    t = t - 1
    E[s][t] = 1
    E[t][s] = 1

# print('edge=', E)


def count(path, E):
    if len(path) == N:
        return 1
    else:
        edges = []
        cur = path[-1]
        ret = 0
        for n in range(N):
            if E[cur][n] == 1 and n not in path:
                npath = path + [n]
                ret += count(npath, E)
        return ret

ret = count([0], E)
print(ret)