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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

// myfirstatcoder.cpp : アプリケーションのエントリ ポイントを定義します。

#include "bits/stdc++.h"
#define rep(i,cc,n) for(int i=cc;i<(n);++i)
#define drep(i,cc,n) for(int i=cc;i>(n);--i)
#define llrep(i,cc,n) for(long long i=cc;i<(n);++i)
using namespace std;
using ll = long long;

int Permutation(int N, vector<int> a, vector<int> b)
{
	int count = 0;

	// one_case = {0,1,2,3,...n-1} とする
	vector<int> node(N);
	iota(node.begin(), node.end(), 0);

	// 頂点と辺の探索 → 辺の元：a[i]  辺の先：b[i]
	vector<vector<int>> list(N, vector<int>(N));
	for (int i = 0;i < a.size();++i)
	{
		--a[i];
		--b[i];
		list[a[i]][b[i]] = 1;
		list[b[i]][a[i]] = 1;
	}

	do // 順列の最後になるまで one_case を並び替えながらループ O(N!)
	{
		int load = 1;
		rep(i,1,N)
			if (list[node[i - 1]][node[i]] == 0)
				load = 0;
		if (load)
			++count;
	} while (next_permutation(node.begin() + 1, node.end()));
	return count;
}

int main()
{
	int N,M;
	cin >> N >> M;
	vector<int> a(M), b(M);
	vector<vector<int>> side(M);
	rep(i, 0, M)
	{
		cin >> a[i] >> b[i];
	}

	cout << Permutation(N,a,b) << endl;
	
	return 0;
}