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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
	"sort"
)

func main() {
	var n, m int
	fmt.Scan(&n, &m)
	g := make([][]int, n)
	for i := range g {
		g[i] = make([]int, n)
	}
	for i := 0; i < m; i++ {
		var a, b int
		fmt.Scan(&a, &b)
		a--
		b--
		g[a][b] = 1
		g[b][a] = 1
	}

	o := make([]int, n-1)
	for i := range o {
		o[i] = i + 1
	}

	var cnt int
	for {
		ok := true
		i := 0
		for _, v := range o {
			if g[i][v] != 1 {
				ok = false
				break
			}
			i = v
		}
		if ok {
			cnt++
		}

		if !nextPermutation(sort.IntSlice(o)) {
			break
		}
	}
	fmt.Println(cnt)
}

func nextPermutation(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}