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<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<functional>
#include<map>
#include<iomanip>
#include<limits>
#include<unordered_set>
#include<cmath>
#include <numeric>
#include <array>
#include<utility>
#include <complex>
#define M_PI 3.141592653589793238
using namespace std;
//long long p = 998244353;
long long p = 1000000007;
#define upperbound(v,val) upper_bound(v.begin(),v.end(),val)-v.begin()
#define lowerbound(v,val) lower_bound(v.begin(),v.end(),val)-v.begin()
#define int long long
#define ll long long
#define vel vector<ll>
#define vvel vector<vel>
#define rep(i,n) for(int i=0;i<n;i++)
#define sor(v) sort(v.begin(),v.end())
#define mmax(a,b) a=max(a,b)
#define mmin(a,b) a=min(a,b)
#define mkp(a,b) make_pair(a,b)
#define pin pair<ll,ll>
#define qin pair<pin,int>
#define V vector
#define Endl endl
#define veb vector<bool>
#define fcout cout << fixed << setprecision(15)
#define rev(s) reverse(s.begin(),s.end())
#define lower(h,val) (lower_bound(h.begin(),h.end(),val)-h.begin())
#define upper(h,val) (upper_bound(h.begin(),h.end(),val)-h.begin())
#define vveb V<veb>
#define omajinai cin.tie(0);ios::sync_with_stdio(false);
#define endl "\n"
#define pb push_back
vel kai;
vel inv_kai;
int root(int x, vel& pa) {
	if (pa[x] == -1) { return x; }
	int ans = root(pa[x], pa); pa[x] = ans;
	return ans;
}
bool mar(int x, int y, vel& pa) {
	x = root(x, pa);
	y = root(y, pa);
	if (x != y) { pa[x] = y; }
	return (x != y);
}
int gcd(int x, int y) {
	if (x < y) { return gcd(y, x); }
	if (y == 0) { return x; }
	return gcd(y, x % y);
}
int lcm(ll x, ll y) {
	x = abs(x); y = abs(y);
	return x * (y / gcd(x, y));
}
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
void make_kai(int max_kai,int p) {
	kai = vel(max_kai, 1);
	inv_kai = kai;
	rep(i, max_kai - 1) {
		kai[i + 1] = kai[i] * (i + 1); kai[i + 1] %= p;
		inv_kai[i + 1] = modinv(kai[i + 1], p);
	}
}
int com(int n, int r) {
	if ((n < 0) || (r < 0) || (r > n)) { return 0; }
	int ans = (kai[n] * inv_kai[r]) % p;
	return (ans * inv_kai[n - r]) % p;
}
vel uni(vel x) {
	if (x.size() == 0) { return x; }
	sor(x);
	int n = x.size();
	vel ans(1, x[0]);
	for (int j = 1; j < n; j++) {
		if (x[j - 1] != x[j]) { ans.push_back(x[j]); }
	}
	x = ans;
	return x;
}
vel dijk(V<V<pin>>& way, int st, int inf) {
	int n = way.size();
	vel dist(n, inf); dist[st] = 0;
	priority_queue<pin, vector<pin>, greater<pin>> pq;
	pq.push(mkp(0, st));
	veb is_checked(n, false);
	while (!pq.empty()) {
		pin x = pq.top(); pq.pop();
		int pot = x.second;
		if (!is_checked[pot]) {
			is_checked[pot] = true;
			for (auto y : way[pot]) {
				int nex_dist = x.first + y.second;
				int nex_pot = y.first;
				if (dist[nex_pot] > nex_dist) {
					dist[nex_pot] = nex_dist;
					pq.push(mkp(nex_dist, y.first));
				}
			}
		}
	}
	return dist;
}
void make_tree(vvel& chi, vel& par,int n) {
	V<V<pin>> way(n);
	rep(i, n - 1) {
		int a, b; cin >> a >> b; a--; b--;
		way[a].push_back(mkp(b, 1));
		way[b].push_back(mkp(a, 1));
	}
	vel dist = dijk(way, 0, n + 1);
	par = vel(n,-1);
	chi = vvel(n);
	rep(i, n) {
		for (auto nex : way[i]) {
			int pot = nex.first;
			if (dist[pot] > dist[i]) { chi[i].push_back(pot); }
			else { par[i] = pot; }
		}
	}
}
void pri(vel& v) {
	if (v.size() == 0) { return; }
	cout << v[0];
	rep(i, v.size() - 1) { cout << " " << v[i + 1]; }
	cout << endl;
	return;
}
int modpow(int a, int n, int p) {
	if (n == 0) { return 1; }
	int m = n / 2;
	int x = modpow(a, n / 2, p);
	x *= x; x %= p;
	if (n % 2 == 1) { x *= a; x %= p; }
	return x;
}
vvel ans;
void solve(int i, string &s) {
	if (ans[0][i] == -1) {
		if (s[i] == 'R') {
			if (s[i + 1] == 'L') {
				ans[0][i] = i;
				ans[1][i] = i + 1;
				ans[0][i + 1] = i + 1;
				ans[1][i + 1] = i;
			}
			else {
				solve(i + 1, s);
				ans[0][i] = ans[1][i + 1];
				ans[1][i] = ans[0][i + 1];
			}
		}
		else {
			solve(i - 1, s);
			ans[0][i] = ans[1][i - 1];
			ans[1][i] = ans[0][i - 1];
		}
	}
}
signed main() {
	int n, m; cin >> n >> m;
	vel jun(n);
	rep(i, n) { jun[i] = i; }
	int ans = 0;
	vvel d(n, vel(n, 0));
	rep(i, m) { int a, b; cin >> a >> b; d[a - 1][b - 1] = 1; d[b - 1][a - 1] = 1; }
	while (jun[0] == 0) {
		bool fl = true;
		rep(i, n - 1) {
			fl &= (d[jun[i]][jun[i + 1]] == 1);
		}
		if (fl) { ans++; }
		next_permutation(jun.begin(), jun.end());
	}
	cout << ans << endl;
	return 0;
}