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 OCaml (4.02.3) (judge verdict: Accepted)

let n, m = Scanf.scanf "%d %d\n" (fun n m -> n, m)
let edge = Array.make (n + 1) []

let rec load_graph i =
  if i < m then
    Scanf.scanf "%d %d\n" (fun a b ->
      edge.(a) <- b :: edge.(a);
      edge.(b) <- a :: edge.(b);
      load_graph (i + 1)
    )

let rec ans v exists rest =
  if rest = 0 then 1
  else
    let rec f ret = function
        [] -> ret
      | u :: l ->
        if exists.(u) then (
          let copy = Array.init (n + 1) (fun i -> exists.(i)) in
          copy.(u) <- false;
          f (ret + ans u copy (rest - 1)) l
        ) else f ret l
    in f 0 edge.(v)

let () =
  load_graph 0;
  let exists = Array.make (n + 1) true in
  exists.(1) <- false;
  print_int (ans 1 exists (n - 1))