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 Ruby (2.3.3) (judge verdict: Runtime Error)

M, N = gets.chomp.split.map(&:to_i)

go = []
(N+1).times do |i|
  go[i] = []
end

M.times do |i|
  a, b = gets.chomp.split.map(&:to_i)
  go[a].push(b)
  go[b].push(a)
end

memo = []
(N+1).times do |i|
  memo[i] = Array.new(N+1, 0)
end

count = 0
(2..N).to_a.permutation(N-1) do |a|
  a.unshift(1)
  flag = false
  (N-1).times do |i|
    if !memo[a[i]][a[i+1]] || !go[a[i]].include?(a[i+1])
      flag = true
      break
    elsif memo[a[i]][a[i+1]] == 0
      memo[a[i]][a[i+1]] == true
    end
  end
  count += 1 unless flag
end
p count