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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val (n, m) = readLine()!!.split(' ').map(String::toInt)
    val g = Array(n) { BooleanArray(n) }
    repeat(m) {
        val (a, b) = readLine()!!.split(' ').map(String::toInt)
        g[a - 1][b - 1] = true
        g[b - 1][a - 1] = true
    }
    val visited = BooleanArray(n)
    fun dfs(v: Int): Int {
        if (visited.all { it }) return 1
        var cnt = 0
        for (i in 0 until n) {
            // println("g[$v][$i] = ${g[v][i]}; visited[$i] = ${visited[i]}")
            if (!g[v][i] || visited[i]) continue
            visited[i] = true
            cnt += dfs(i)
            visited[i] = false
        }
        return cnt
    }
    visited[0] = true
    println(dfs(0))
}