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 <bits/stdc++.h>

using namespace std;
using ll = long long;

using Field = vector<vector<int>>;
using Graph = vector<vector<int>>;
using VI = vector<int>;
using VC = vector<char>;
using PI = pair<int, int>;

#define REP(i, n) for (int i = 0; i < (n); i++)
#define ALL(x) x.begin(), x.end()


int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    
    int n, m; cin >> n >> m;
    Graph G(n, vector<int>());

    REP(i, m) {
        int a, b; cin >> a >> b;
        a--, b--;
        G.at(a).push_back(b);
        G.at(b).push_back(a);
    }

    VI np;
    REP(i, n) np.push_back(i);

    sort(np.begin(), np.end());

    int ans = 0;

    do {
        bool flag = true;
        if (np.at(0) != 0) continue;
        REP(i, n-1) {
            int from = np.at(i);
            int to = np.at(i + 1);
            if (find(G.at(from).begin(), G.at(from).end(), to) == G.at(from).end()){
                flag = false;
                break;
            }
        }
        if (flag) ans++;

    }while(next_permutation(np.begin(), np.end()));

    cout << ans << endl;
    
    return 0;
}