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: Accepted)

n, m = gets.split.map(&:to_i)
graph = Array.new(n+1){Array.new(n+1, false)}
edge = []
m.times do |i|
  edge[i] = gets.split.map(&:to_i)
  graph[edge[i][0]-1][edge[i][1]-1] = graph[edge[i][1]-1][edge[i][0]-1] = true
end
dp = Array.new(1<<n) {Array.new(n,0)}
dp[1][0] = 1
(1<<n).times do |s|
  n.times do |v|
    #vがSに含まれるならskip
    next if (1 << v) & s > 0

    #vがSに含まれないなら，dp[S+v][v]=\sum_u dp[S][u]を計算する
    0.upto(n-1) do |u|
      #和をとる条件：uはSに含まれ，かつvに接続している
      if ((1<<u) & s) > 0 && graph[u][v]
        dp[s|(1<<v)][v] += dp[s][u]
      end
    end
  end
end

puts dp[(1<<n)-1].inject(:+)