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)

#include <cassert>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <vector>
#include <map>
#include <queue>
#include <numeric>
#include <algorithm>

using namespace std;
using lint = long long;
constexpr int MOD = 1000000007, INF = 1111111111;
constexpr lint LINF = 1LL << 60;

template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
	for (const auto &e : vec) os << e << (&e == &vec.back() ? "" : " ");
	return os;
}

#ifdef _DEBUG
template <class T>
void dump(const char* str, T &&h) { cerr << str << " = " << h << "\n"; };
template <class Head, class... Tail>
void dump(const char* str, Head &&h, Tail &&... t) {
	while (*str != ',') cerr << *str++; cerr << " = " << h << "\n";
	dump(str + (*(str + 1) == ' ' ? 2 : 1), t...);
}
#define DMP(...) dump(#__VA_ARGS__, __VA_ARGS__)
#else 
#define DMP(...) ((void)0)
#endif

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int N, M;
	cin >> N >> M;

	vector<vector<int>> edge(N, vector<int>(N));
	int u, v;
	for (int i = 0; i < M; i++) {
		cin >> u >> v;
		u--, v--;
		edge[u][v] = 1;
		edge[v][u] = 1;
	}

	vector<int> per(N);
	iota(per.begin(), per.end(), 0);
	int ans = 0;
	do {

		bool flag = true;
		for (int i = 0; i < N - 1; i++) {
			if (!edge[per[i]][per[i + 1]]) {
				flag = false;
				break;
			}
		}
		ans += flag;
		if (flag) DMP(per);
	} while (next_permutation(per.begin() + 1, per.end()));

	cout << ans << "\n";

	return 0;
}