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: Wrong Answer)

n, m = gets.chomp.split.map(&:to_i)

h = []
m.times do |i|
  h[i] = gets.chomp.split.map {|a| a.to_i - 1 }
end

reached = Array.new(n)
reached[0] = true

def search(from, n, m, h, reached)
  ans = 0
  queue = []
  h.each do |a|
    if a[0] == from
      queue.push([from, a[1], reached])
    elsif a[1] == from
      queue.push([from, a[0], reached])
    end
  end
  while !queue.empty?
    from, to, reached = queue.shift
    reached[to] = true
    if reached.all?
      ans += 1
    else
      h.each do |a|
        if a[0] == to && !reached[a[1]]
          queue.push([to, a[1], reached.dup])
        elsif a[1] == to && !reached[a[0]]
          queue.push([to, a[0], reached.dup])
        end
      end
    end
  end
  p ans
end

search(0, n, m, h, reached)