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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
typedef long long int ll;
typedef pair<int, int> ii;
#define DEBUG freopen("in.txt", "r", stdin);

struct fastio {
    fastio() {
        ios::sync_with_stdio(false);
        cout << setprecision(10) << fixed;
        cin.tie(0);
    }
};

fastio _fast_io;

const int N = 8;
bool e[N][N];
int n, m, f, t, ans;
int a[N];

int main() {
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        cin >> f >> t;
        e[f - 1][t - 1] = true;
        e[t - 1][f - 1] = true;
    }
    iota(a, a + n, 0);
    do {
        bool ok = true;
        for (int i = 0; i < n - 1; ++i) {
            if (!e[a[i]][a[i + 1]]) {
                ok = false;
                break;
            }
        }
        if (ok)
            ++ans;
    } while (next_permutation(a + 1, a + n));

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